1. When x is a prime number
ax - a is always divisible by x.
Q1. What is remainder when 5101 - 5 is divided by 101.
(1) 0 (2) 100 (3) 5 (4) 50 (5) 20
Solution :- As concept states, it is divisible by 101. Hence remainder is zero.
Q2. 91101 -91 is divisible by numbers except ?
(1) 91 (2) 13 (3) 7 (4) 101 (5) Divisible by all above
Solution :- As concept states, it is divisible by 101 hence option 5 is eliminated.
91101 -91 = 91(91100 -1), hence divisible by 91
91 = 13 * 7 hence option 5.
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2. When a base of a number increases, the number decreases and vice-versa.
Q3. (2345)6 is equal to
(1) (569)10 (2) (3458)10 (3) (4286)10 (4) (569)5
Solution :- Only option 1 follows concept. No need to solve.
Q3. (2345)6 is equal to
(1) (569)10 (2) (3458)10 (3) (4286)10 (4) (569)5
Solution :- Only option 1 follows concept. No need to solve.
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3. Relation between a two digit number and number formed using its digits.
ab = 10a +b &
ba = 10b + a
hence ab - ba = (10a +b) - (10b + a) = 9(a-b)
& ab + ba = (10a +b) + (10b + a) = 11(a+b)
Q4. The difference between a two digit number and number formed by reversing its digits is 45. What is difference between the digits of its number ?
(1) 9 (2) 5 (3) 3 (4) 4 (5) None of these
Solution :- According to concept 9(a-b) = 45 hence (a-b) 5. Hence option 2.
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4.IMP - Divisible count in a range.
Steps to follow.
I. Subtract extreme numbers & divide by number (x)
II. Check remainder (n)
III. Check first n digits of range, if one of the number is divisible then add 1 to x.
Q5. How many three digit numbers are divisible by 7 ?
(1) 128 (2) 129 (3) 127 (4) 142 (5) 140
Solution :-
Extreme three digit numbers are 100 - 999
I. 999 - 100 = 899 the 899 / 7 = 128 * 7 + 3
II. Remainder = 3
III. None of the first three digits (100,101 & 102) are divisible by 7 hence answer is option 1.
Q6. How many four digit numbers are divisible by 7 ?
(1) 1284 (2) 1285 (3) 1286 (4) 1428 (5) 1287
Solution :-
Extreme four digit numbers are 1000 - 9999
I. 9999 - 1000 = 8999 the 8999 / 7 = 1285 * 7 + 4
II. Remainder =4
III.Out of the first four digits (1000,1001,1002 & 1003) number 1001 is divisible by 7 hence answer = 1285 + 1 = 1286. Hence option 3.
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5. Product of three consecutive numbers is always divisible by 6.
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6. To find highest power of a number in factorial .
I. Factorize the number.
II. Find the power of Individual factor in factorial by simply dividing number continuously with ignoring remainder. Add all the numbers.
Q7. Find highest power of 15 in 30!
(1) 2 (2) 6 (3) 21 (4) 7 (5) 14
Solution: Prime Factors of 15 = 3 * 5.
To find highest power of 15 just find highest power of 5.
Highest power of 5 = 30/5 + 6/5 (ignore remainders for next calculation)
= 6 + 1 = 7.
As proved highest power of 15 in 30! is 7.Hence answer option 4.
Q8. Find highest power of 12 in 30!.
(1) 3 (2) 26 (3) 14 (4) 13 (5) 12
Solution: Prime Factors of 12 = 22 * 3.
Highest power of 2 = 30/2 + 15/2 + 7/2 + 3/2 (ignore remainders for next calculation)
=15 + 7 + 3 + 1 = 26 => 13 pairs of 2.
Highest power of 22 = 13.
Highest power of 3 = 30/3 + 9/3 + 3/3 (ignore remainders for next calculation)
= 10 + 3 + 1 = 14.
As proved highest power of 12 in 30! is 13.Hence answer option 4.
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7. Sum of numbers of numbers formed using n digits.
Sum = Sum of digits * (n-1)! * 1111.... n times.
provided no digit is zero & digits are not repeated.
Q9. Find sum of all digits formed using 2,3,4 & 6 such then digits are not repeated.
(1) 99990 (2) 62525 (3) 99900 (4) 98900 (5) None of these.
Solution :-
Since zero is not one of digits & digits are not repeated, we can use formula
Sum = Sum of digits * (n-1)! * 1111.... n times.
sum = 15 * 3!* 1111 = 99990. Hence option 1.
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8. All numbers which are perfect squares of odd numbers are of the form 8k+1 but vice-versa not true.
OR a perfect square of a odd number leaves a remainder of 1 when divided by 8.
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9. A four digit number formed with its 2 digits repeating (abab) = ab * 101
A six digit number formed with its 3 digits repeating (abcabc) = abc * 1001 = abc * 11 * 13 * 7
(Very frequently asked question pattern in any exams. Concept saves a lot of time)
Q10. Number 2828 is divisible by except
(1) 2 (2) 7 (3) 14 (4) 101 (5) 11
Solution :- I know your reaction would be "Wow i can answer this without calculations, answer is option 5.
Q11. Number 123123 is divisible by except
I. 123
II. 11
III. 77
IV. 143
V. 91
(1) Statement I, II but not III, IV.
(2) Statement II, III but not I.
(3) Statement I, II, IV but not III.
(4) Statement I, II, III, IV but not V.
(5) Divisible by all
Solution :
123123 = 123 * 1001 = 123 * 7 * 11 * 13 = 123 * 7 * 143 = 123 * 77 * 13 = 123 * 91 * 11
hence answer option 5.
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10.Number of Factors of a number is determined by power of its prime factors.
Number form in prime factors = ax * by * cz ....
where a, b , c ... are prime numbers. then
Number of factors = (x+1)(y+1)(z+1)....
Q12. Find the number of factors of number 96 ?
Solution: N = 96 = 25 *31
Number of factors = (5+1)(1+1) = 6 * 2 = 12
Note:
1. In a perfect square power of prime factors are always even and they always have odd number of factors.
2. In a perfect cube power of prime factors is always multiple of three.
Q13. How three many numbers with odd number of factors are divisible by 3 ?
(1) 7 (2) 21 (3) 149 (4) 150 (5) 299
Solution : Only perfect squares have odd number of factors. And three digit perfect squares divisible by 3 are 144, 225, 324, 441, 576, 729 & 900. Hence option 1.
Q14. Which of the following can not be number of factors of a number which is a perfect cube ?
(1) 4 (2) 16 (3) 7 (4) 28 (5)9
Solution : Power of prime factors of a perfect cube are always multiple of 3.
3. Relation between a two digit number and number formed using its digits.
ab = 10a +b &
ba = 10b + a
hence ab - ba = (10a +b) - (10b + a) = 9(a-b)
& ab + ba = (10a +b) + (10b + a) = 11(a+b)
Q4. The difference between a two digit number and number formed by reversing its digits is 45. What is difference between the digits of its number ?
(1) 9 (2) 5 (3) 3 (4) 4 (5) None of these
Solution :- According to concept 9(a-b) = 45 hence (a-b) 5. Hence option 2.
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4.IMP - Divisible count in a range.
Steps to follow.
I. Subtract extreme numbers & divide by number (x)
II. Check remainder (n)
III. Check first n digits of range, if one of the number is divisible then add 1 to x.
Q5. How many three digit numbers are divisible by 7 ?
(1) 128 (2) 129 (3) 127 (4) 142 (5) 140
Solution :-
Extreme three digit numbers are 100 - 999
I. 999 - 100 = 899 the 899 / 7 = 128 * 7 + 3
II. Remainder = 3
III. None of the first three digits (100,101 & 102) are divisible by 7 hence answer is option 1.
Q6. How many four digit numbers are divisible by 7 ?
(1) 1284 (2) 1285 (3) 1286 (4) 1428 (5) 1287
Solution :-
Extreme four digit numbers are 1000 - 9999
I. 9999 - 1000 = 8999 the 8999 / 7 = 1285 * 7 + 4
II. Remainder =4
III.Out of the first four digits (1000,1001,1002 & 1003) number 1001 is divisible by 7 hence answer = 1285 + 1 = 1286. Hence option 3.
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5. Product of three consecutive numbers is always divisible by 6.
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6. To find highest power of a number in factorial .
I. Factorize the number.
II. Find the power of Individual factor in factorial by simply dividing number continuously with ignoring remainder. Add all the numbers.
Q7. Find highest power of 15 in 30!
(1) 2 (2) 6 (3) 21 (4) 7 (5) 14
Solution: Prime Factors of 15 = 3 * 5.
To find highest power of 15 just find highest power of 5.
Highest power of 5 = 30/5 + 6/5 (ignore remainders for next calculation)
= 6 + 1 = 7.
As proved highest power of 15 in 30! is 7.Hence answer option 4.
Q8. Find highest power of 12 in 30!.
(1) 3 (2) 26 (3) 14 (4) 13 (5) 12
Solution: Prime Factors of 12 = 22 * 3.
Highest power of 2 = 30/2 + 15/2 + 7/2 + 3/2 (ignore remainders for next calculation)
=15 + 7 + 3 + 1 = 26 => 13 pairs of 2.
Highest power of 22 = 13.
Highest power of 3 = 30/3 + 9/3 + 3/3 (ignore remainders for next calculation)
= 10 + 3 + 1 = 14.
As proved highest power of 12 in 30! is 13.Hence answer option 4.
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7. Sum of numbers of numbers formed using n digits.
Sum = Sum of digits * (n-1)! * 1111.... n times.
provided no digit is zero & digits are not repeated.
Q9. Find sum of all digits formed using 2,3,4 & 6 such then digits are not repeated.
(1) 99990 (2) 62525 (3) 99900 (4) 98900 (5) None of these.
Solution :-
Since zero is not one of digits & digits are not repeated, we can use formula
Sum = Sum of digits * (n-1)! * 1111.... n times.
sum = 15 * 3!* 1111 = 99990. Hence option 1.
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8. All numbers which are perfect squares of odd numbers are of the form 8k+1 but vice-versa not true.
OR a perfect square of a odd number leaves a remainder of 1 when divided by 8.
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9. A four digit number formed with its 2 digits repeating (abab) = ab * 101
A six digit number formed with its 3 digits repeating (abcabc) = abc * 1001 = abc * 11 * 13 * 7
(Very frequently asked question pattern in any exams. Concept saves a lot of time)
Q10. Number 2828 is divisible by except
(1) 2 (2) 7 (3) 14 (4) 101 (5) 11
Solution :- I know your reaction would be "Wow i can answer this without calculations, answer is option 5.
Q11. Number 123123 is divisible by except
I. 123
II. 11
III. 77
IV. 143
V. 91
(1) Statement I, II but not III, IV.
(2) Statement II, III but not I.
(3) Statement I, II, IV but not III.
(4) Statement I, II, III, IV but not V.
(5) Divisible by all
Solution :
123123 = 123 * 1001 = 123 * 7 * 11 * 13 = 123 * 7 * 143 = 123 * 77 * 13 = 123 * 91 * 11
hence answer option 5.
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10.Number of Factors of a number is determined by power of its prime factors.
Number form in prime factors = ax * by * cz ....
where a, b , c ... are prime numbers. then
Number of factors = (x+1)(y+1)(z+1)....
Q12. Find the number of factors of number 96 ?
Solution: N = 96 = 25 *31
Number of factors = (5+1)(1+1) = 6 * 2 = 12
Note:
1. In a perfect square power of prime factors are always even and they always have odd number of factors.
2. In a perfect cube power of prime factors is always multiple of three.
Q13. How three many numbers with odd number of factors are divisible by 3 ?
(1) 7 (2) 21 (3) 149 (4) 150 (5) 299
Solution : Only perfect squares have odd number of factors. And three digit perfect squares divisible by 3 are 144, 225, 324, 441, 576, 729 & 900. Hence option 1.
Q14. Which of the following can not be number of factors of a number which is a perfect cube ?
(1) 4 (2) 16 (3) 7 (4) 28 (5)9
Solution : Power of prime factors of a perfect cube are always multiple of 3.
let a & b any prime factors. then
a3 then number of factors = 3 + 1 = 4
a6 then number of factors = 6 + 1 = 7
a3 * b3 then number of factors = 4 * 4 = 16
a6 * b3 then number of factors = 7 * 4 = 28
hence option 5.
Or We knows that only perfect square can have odd number of factors so now we are only left with option 3 & 5. We can easily find the number a6 is a perfect cube as well as perfect square. Hence option 5.
Q15. Find the number of factors in 66 - 56
(1) 4 (2) 8 (3) 16 (4) 36 (5) 49
Solution:
66 - 56 = (63)2- (53)2
= (216)2- (125)2 = 91 * 341 = 13 * 7 * 31 * 11
Number of factors = 2 * 2 * 2 * 2 = 16. Hence answer option 3.
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11. Number of ways in which a number can be written as a product of two co-prime numbers = 2M-1. where M is number of prime factors.
Q16. In how many number of ways 540 can be written as a product of two co-prime numbers ?
(1) 1 (2) 2 (3) 3 (4) 4 (5) 6
Solution:- 540 = 22 * 33 * 51
Number of ways in which a number can be written as a product of two co-prime numbers
= 2M-1 = 2M-1 = 23-1 = 4. Hence option 4.
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12. A perfect square will never end in 2,3,7 &8.
CET & SNAP Pattern
Q17. Which of the following is not a perfect square ?
(1) 2025 (2) 3844 (3) 6568 (4) 7056 (5) 3025
Solution: - As concept states answer is option 3. ( Don't try to calculate in exams.)
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13. A square of n digit number will have either 2n or 2n-1 digits.
Q18. Which of the following number , with which some of digits expressed as 'x' can be a perfect square of 5 digit number ?
(1) 30xxxxxxxxx8
(2) 30xxxx25
(3) 4xxxxxx6
(4) 1xxxxxxx0
(5) 4xxxxx25
Sol:- Square of a 5 digit number will have 9 digits or 10 digits. Hence option 2, 3 & 5 are eliminated.
A square can not end with 8 so option 1 is eliminated. Hence option 4 is answer.
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14. Series type.
I. Sum of first 'n' natural numbers = n(n+1) / 2
II. Sum of first 'n' odd numbers = n2
III. Sum of first 'n' even numbers = n2 + n
IV. Sum of the square of first 'n' natural numbers = n(n+1)(2n+1)/6
V. Sum of the cube of first 'n' natural numbers = (n(n+1) / 2)2
Q19. What is sum of all three digit numbers lesser than 251 ?
(1) 4950 (2) 14825 (3) 26250 (4) 26425 (5) 31375
Sol:-
Sum of all digits from 1 - 250(Series formula I) = 250 * 251 / 2 = 31375
Sum of all digits from 1-99 = 99 * 100 / 2 = 4950
Sum of all three digits numbers less than 251 i.e. 100-250 = 31375 - 4950 = 26425
Hence answer option 4.
Or
Total number between 100 - 250 = 151
Avg of numbers = 175
sum of numbers = 151 * 175 = 26425
Q20. What is sum of all numbers which are perfect cube and less than 8001 ?
(1) 21000 (2) 32600 (3) 36800 (4) 42100 (5) 44100
a3 * b3 then number of factors = 4 * 4 = 16
a6 * b3 then number of factors = 7 * 4 = 28
hence option 5.
Or We knows that only perfect square can have odd number of factors so now we are only left with option 3 & 5. We can easily find the number a6 is a perfect cube as well as perfect square. Hence option 5.
Q15. Find the number of factors in 66 - 56
(1) 4 (2) 8 (3) 16 (4) 36 (5) 49
Solution:
66 - 56 = (63)2- (53)2
= (216)2- (125)2 = 91 * 341 = 13 * 7 * 31 * 11
Number of factors = 2 * 2 * 2 * 2 = 16. Hence answer option 3.
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11. Number of ways in which a number can be written as a product of two co-prime numbers = 2M-1. where M is number of prime factors.
Q16. In how many number of ways 540 can be written as a product of two co-prime numbers ?
(1) 1 (2) 2 (3) 3 (4) 4 (5) 6
Solution:- 540 = 22 * 33 * 51
Number of ways in which a number can be written as a product of two co-prime numbers
= 2M-1 = 2M-1 = 23-1 = 4. Hence option 4.
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12. A perfect square will never end in 2,3,7 &8.
CET & SNAP Pattern
Q17. Which of the following is not a perfect square ?
(1) 2025 (2) 3844 (3) 6568 (4) 7056 (5) 3025
Solution: - As concept states answer is option 3. ( Don't try to calculate in exams.)
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13. A square of n digit number will have either 2n or 2n-1 digits.
Q18. Which of the following number , with which some of digits expressed as 'x' can be a perfect square of 5 digit number ?
(1) 30xxxxxxxxx8
(2) 30xxxx25
(3) 4xxxxxx6
(4) 1xxxxxxx0
(5) 4xxxxx25
Sol:- Square of a 5 digit number will have 9 digits or 10 digits. Hence option 2, 3 & 5 are eliminated.
A square can not end with 8 so option 1 is eliminated. Hence option 4 is answer.
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14. Series type.
I. Sum of first 'n' natural numbers = n(n+1) / 2
II. Sum of first 'n' odd numbers = n2
III. Sum of first 'n' even numbers = n2 + n
IV. Sum of the square of first 'n' natural numbers = n(n+1)(2n+1)/6
V. Sum of the cube of first 'n' natural numbers = (n(n+1) / 2)2
Q19. What is sum of all three digit numbers lesser than 251 ?
(1) 4950 (2) 14825 (3) 26250 (4) 26425 (5) 31375
Sol:-
Sum of all digits from 1 - 250(Series formula I) = 250 * 251 / 2 = 31375
Sum of all digits from 1-99 = 99 * 100 / 2 = 4950
Sum of all three digits numbers less than 251 i.e. 100-250 = 31375 - 4950 = 26425
Hence answer option 4.
Or
Total number between 100 - 250 = 151
Avg of numbers = 175
sum of numbers = 151 * 175 = 26425
Q20. What is sum of all numbers which are perfect cube and less than 8001 ?
(1) 21000 (2) 32600 (3) 36800 (4) 42100 (5) 44100
Solution :-
8000 is cube of 20. Hence question is asking of sum of all first 20 cubes. Which can be calculated using 5th formula.
sum = (20(20+1) / 2)2
= (20(21) / 2)2
= 210 * 210
= 44100. Hence answer option 5.
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Go back to concepts to learn more-> click here
8000 is cube of 20. Hence question is asking of sum of all first 20 cubes. Which can be calculated using 5th formula.
sum = (20(20+1) / 2)2
= (20(21) / 2)2
= 210 * 210
= 44100. Hence answer option 5.
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Go back to concepts to learn more-> click here
This is so far my best posting. Very very important for any entrance exam.
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