Examples
Example 1. Find the largest number which divides 238, 358 and 418 leaving the same remainder in each case.(A) 60 (B) 120 (C) 45 (D) 90
Solution:
Let the largest number which divides 238, 358 and 418 and leaves the same remainder (say r) n each case be N.
238 = Na + r --- (1)
358 = Nb + r --- (2)
418 = Nc + r --- (3)
a, b and c are the respective quotients when 238, 358 and 418 are divided by N.
Subtracting equation (1) from (2), 120 = N (b - a)
Subtracting equation (2) from (3), 60 = N (c - b)
As N is the largest possible number it must be the HCF (60, 120) i.e 60. Answer: A
Example 2. Which of the following is the smallest five-digit number which when divide by 8, 11 and 24 leaves a remainder of 5 in each case?
(A) 10125 (B) 10037 (C) 10301 (D) 10061
Solution:
Let the smallest five-digit number be N.
N = k [(LCM of (8,11,24)] + 5, where k is an integer
N = 264k + 5
Each option gives us the value of N.
Subtracting 5 from each option, the smallest result which is divisible by 264 must be the result.
As 10037 - 5 = 10032 is the only multiple of 264, this is the smallest five-digit number satisfying the condition. Answer: B
Example 3. Find the sum of the remainders obtained when a number n is divided by 9 and 7 successively given that n is the smallest number that leaves remainders of 4, 6 and 9 when divided successively by 13, 11 and 15.
(A) 5 (B) 6 (C) 4 (D) 9
Solution:
The smallest number 'n' which when divided by 13, 11 and 15 successively leaves remainders of 4, 6 and 9 is obtained as shown below:
13 11 15
4 6 9
As shown, the divisors are in the order of which division is carried out and their respective remainder is written below.
Starting from the last remainder, we go diagonally left upwards to the first row multiplying and then directly below adding the figure already obtained. We continue this process till we reach the figure on the extreme left in the second row.
We start at the bottom right corner 9, and go from 2nd row to 1st row diagonally to the left multiplying. We get 9 x 11 = 99, then we come down to the 2nd row adding, we get 99 + 6 = 105. Again, multiplying diagonally left upwards, we get 105 x 13 = 1365. Coming down to 2nd row, adding, we get 1365 + 4 = 1369.
The quotient and remainder when 1369 is divided by 9 are 152 and 1 respectively.
The quotient and remainder when 152 is divided by 7 are 21 and 5 respectively.
Sum of remainders = 1 + 5 = 6. Answer: B
Example 4. Find the unit's digit of 34563²⁰³⁵⁹ + 2358⁷⁸⁴
(A) 3 (B) 4 (C) 5 (D) 6
Solution:
Last digit of 34563²⁰³⁵⁹ = last digit of 3²⁰³⁵⁹ = last digit of 3³ = 7
(3 to the power any number has unit's repetition cycle of 4. 20359 is of the form 4k + 3)
Last digit of 2358⁷⁸⁴ = last digit of 8⁷⁸⁴ = last digit of 8⁴ = 6
(8 to the power any number has unit's repetition cycle of 4. 784 is of the form 4k + 4)
∴ The last digit of (34563²⁰³⁵⁹ + 2358⁷⁸⁴) = last digit of (7 + 6) = last digit of 13 = 3
Answer: A
Example 5. Find the remainder of 3⁴⁰ divided by 11.
(A) 9 (B) 1 (C) 3 (D) 5
Solution:
Let us try to find the remainder using pattern method.
Rem (3¹ / 11) = 3
Rem (3² / 11) = 9
Rem (3³ / 11) = 5
Rem (3⁴ / 11) = 4
Rem (3⁵ / 11) = 1
Rem (3⁶ / 11) = 3
Rem (3⁷ / 11) = 9.
Hence the pattern is found to repeat after a cycle of 5.
40 can be written as (5 x 7 + 5)
Hence, the remainder of (3⁴⁰ / 11) = Rem (3⁵ / 11) = 1
Answer: B
Example 6. Find the largest number which divides 606, 732 and 915 leaving remainders of 6, 12 and 15 respectively.
(A) 75 (B) 120 (C) 60 (D) 90
Solution:
Let the largest number satisfying the condition be N.
606 = Na + 6 --- (1)
732 = Nb + 12 --- (2)
915 = Nc + 15 --- (3)
In the equations (1), (2) and (3), a, b and c are the respective quotients when N divides 606, 732 and 915.
606 - 6 = 600 = Na
732 - 12 = 720 = Nb
915 - 15 = 900 = Nc
As N is the largest possible number satisfying the above three equations.
N = HCF (600,720,900) = 60. Answer: C
Example 7. Find the number of zeroes at the end of 175!
(A) 33 (B) 53 (C) 23 (D) 43
Solution:
A zero is formed by the combination of a 5 and a 2.
Number of zeros at the end of 175! is the largest power of 10 in 175!.
As 5 > 2, largest power of 5 in 175! < largest power of 2 in 175!
Hence, the largest power of 5 equals largest power of 10 in 175!
Largest power of 5 in 175! is found as shown below (since 5 is a prime number, we can use the below method)
5) 175 → Given number
5) 35 → Quotient
5) 7 → Quotient
5) 1 → Quotient
Since we can't divide the quotient any more, we stop here.
Add all the quotients = 35 + 7 + 1 = 43.
Hence, the number of zeros at the end of 175! is 43. Answer: D
Example 8. 17n + 7n is always divisible by which of the following if n is odd?
(A) 17 (B) 11 (C) 13 (D) 3
Solution:
an + bn will always be divisible by a + b if n is odd.
As n is odd, always be divisible by 17 + 7 i.e. 24.
As 17n + 7n is divisible by 24, it is also divisible by factors of 24. Of the options, 3 is the only factor of 24.
Hence, 17n + 7n is always divisible by 3. Answer: D
Example 9. Find the unit's digit of the remainder of 59n - 31n divided by 28.
(A) 14 (B) 0 (C) 21 (D) 7
Solution:
an - bn will always be divisible by a - b.
Hence, 59n - 31n will be divisible by 59 - 31 = 28.
Hence, unit's digit of the remainder of 59n - 31n divided by 28 is 0. Answer: B
Example 10. Find the sum of all the numbers which are co-prime to 60 and which are less than 60.
(A) 300 (B) 480 (C) 270 (D) 360
Solution:
If N is a number that can be written as ap . bq . cr...... then, the number of co-primes of N, which are less than N, is:
N(1 - 1/a)(1 - 1/b)(1 - 1/c)......
60 = 2² x 3 x 5
Number of numbers co-prime to 60 = 60(1 - 1/2)(1 - 1/3)(1 - 1/5) = 16
Sum of co-primes of N, that are less than N is N/2 x number of co-primes less than N
= 60/2 x 16 = 480. Answer: 480
Example 11. The remain (17³ + 19³ + 21³ + 23³) divided by 80 is
(A) 20 (B) 60 (C) 0 (D) 40
Solution:
17³ + 19³ + 21³ + 23³ = (17³ + 23³) + (19³ + 21³)
= (17 + 23) [17² + 17(23) + 23²] + (19 + 21) [19² + 19(21) + 21²]
= 40 [17² + 19² + 21² + 23² + 17.23 + 19.21]
The bracket is the sum of 6 odd numbers, i.e. an even number. Therefore, the given expression is a multiple of 80 and hence the remainder when divided by 80 is 0. Answer: C
Example 12. A rectangular floor of dimensions 3 m 60 cm and 5 m 40 cm is to be covered with identical square tiles. Find the minimum number of square tiles required.
(A) 12 (B) 60 (C) 6 (D) 24
Solution:
Let the minimum number of identical square tiles, of side x cm be N.
Area of each square tile = x²
N = Area of rectangular floor / x² = 360/x * 540/x
As the room must be perfectly covered with square tiles, x must be a factor of both 360 and 540. Hence, x = HCF (360, 540) = 180 in order for N to be minimum.
Hence, N = 360/180 * 540/180 = 6. Answer: C
Example 13. Find the largest power of 24 which divides 150!
(A) 38 (B) 48 (C) 54 (D) 60
Solution:
24 = 2³ x 3
Largest power of 2 which divides 150! is found as shown below: (Divide the quotients by 2 till we can no longer divide (less than 2)
2) 150
2) 75 → Quotient
2) 37
2) 18
2) 9
2) 4
2) 2
2) 1
Sum of all the quotients = 75 + 37 + 18 + 9 + 4 + 2 + 1 = 146
Largest power of 2 which divides 150! is 146. Largest power of 2³ which divides 150! will be the quotient of 146/3 i.e. 48
Largest power of 3 can be found to that shown in a way similar to be 72 (= 50 + 16 + 5 + 1).
As we have 48 - 8s and 72 - 3s, we can only combine 48 - 8s with 48 - 3s to form 48 - 24s.
Hence, the largest power of 24 must be 48 [i.e. min(48, 72)] Answer: B
Example 14. For any prime number p greater than 3, p² - 1 is always divisible by
(A) 24 (B) 48 (C) 18 (D) None of these
Solution:
p² - 1 = (p - 1)(p + 1)
The only even prime number is 2.
As p > 3, it is always odd. Hence p - 1 and p + 1 will both be even and one of them will be multiple of 4.
Of the three consecutive natural numbers (p - 1), p and (p + 1), one of them will always be a multiple of 3.
Hence, p² - 1 must be a multiple of (2) (3) (4) i.e. 24. Answer: A
Example 15. There are three cakes weighing 210 gms, 540 gms and 720 gms respectively. Find the minimum number of pieces into which these cakes can be cut such that all the pieces have equal weight.
(A) 72 (B) 49 (C) 36 (D) 60
Solution:
Let the weight of each cake be w gms
Number of pieces of cake = 210/w + 540/w + 720/w. This is minimum only if w is the largest common factor of 210, 540 and 720 i.e. 30.
Minimum number of pieces = 210/30 + 540/30 + 720/30 = 1470/30 = 49. Answer: B
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