The numbers 400, 536 and 645, when divided by a positive integer N, gi

400modN = 22
536modN = 23
645modN = 24

136modN=1
109modN=1

27modN=0
N = 27k

Let N = 54;
400mod54= 22
536mod54 = -4mod54 = 50
645mod54 = -3mod54 = 51

Largest value of N = 27

IMO D

The numbers 400, 536 and 645, when divided by a positive integer N, give the remainders of 22, 23 and 24 respectively. What is the greatest possible value of N?

A. 9
B. 18
C. 21
D. 27 --> correct
E. 54

Solution:
400 = N*d1 + 22 => N*d1=378 = 3^3*14 ---(i)
536 = N*d2 + 23 => N*d2=513 = 3^3*19---(ii)
645 = N*d3 + 24 => N*d3=621 = 3^3 *23--(iii)

HCF of (i), (ii) & (iii) => 3^3=27

Solution

Given
In this question, we are given that

• The numbers 400, 536 and 645, when divided by a positive integer N, give the remainders of 22, 23 and 24 respectively

To find
We need to determine

• The greatest possible value of N

Approach and Working out
When 400 is divided by N, the remainder is 22

• Hence, 400 – 22 = 378 should be completely divisible by N

When 536 is divided by N, the remainder is 23

• Hence, 536 – 23 = 513 should be completely divisible by N

When 645 is divided by N, the remainder is 24

• Hence, 645 – 24 = 621 should be completely divisible by N

Now, the greatest possible value of N = the greatest number that divides 378, 513, 621 = GCD (378, 513, 621) = 27

Thus, option D is the correct answer.

Correct Answer: Option D

easiest way to solve this problem is by substracting remainder from given number for example do 400-22 =388 and then match from the options.

400 = pN+22 => 378 = pN (p is quotient)
536 = qN+23 => 513 = qN (q is quotient)
645 = rN+24 => 621 = rN (r is quotient)

378 = 2 * 3^3 * 7
513 = 3^3 * 19
621 = 3^3 * 23

So common factor dividing all the three numbers (378,513, and 621) is 3^3 = 27

Answer D

400 = pN+22 => 378 = pN (p is quotient)
536 = qN+23 => 513 = qN (q is quotient)
645 = rN+24 => 621 = rN (r is quotient)

378 = 2 * 3^3 * 7
513 = 3^3 * 19
621 = 3^3 * 23

So the common factor dividing all the three numbers (378,513, and 621) is 3^3 = 27

Answer D

To add my inputs to above solutions, remainder is always less than divisor. So options A to C are out straight away.

Between D and E, we may use the approach mentioned above.

Look at the answer choices. This is a variation of PITA (Plugging In The Answers)
A. If we were dividing by 9, how on earth would we end up with a remainder of 22?!? 22 would just mean that we have two more multiples of 9 and then a remainder of 4. A is wrong.
B. If we were dividing by 18, how on earth would we end up with a remainder of 23?!? 23 would just mean we have another multiple of 18 and then a remainder of 5. B is wrong.
C. If we were dividing by 21, how on earth would we end up with a remainder of 24?!? 24 would just mean we have another multiple of 21 and then a remainder of 3. C is wrong.
D. I don't know. Leave it.
E. If we're dividing 645 (an odd number) by 54 (an even number), we'd end up with a remainder that's odd. 24 is not odd. E is wrong.

Answer choice D.


ThatDudeKnowsPITA

The numbers 400, 536 and 645, when divided by a positive integer N, give the remainders of 22, 23 and 24 respectively. What is the greatest possible value of N?

A. 9
B. 18
C. 21
D. 27
E. 54

The reminder cannot be higher than the dividend so A, B and C are wrong. Try to use the solution D and E but only 27 works so the answer is D.
You can do it in less than 30 seconds and reducing the operation done