When positive integer n is divided by 25, the remainder is 13. What is the value of n?

1) N < 100
2) When N is divided by 20, the remainder is 3

C*

How should I think about remainders? I find these types of problems confusing.

When positive integer n is divided by 25, the remainder is 13. What is the value of n?

1) N < 100
2) When N is divided by 20, the remainder is 3



N/25=x+13 --> N= 25x+13 ---(Given)

What is N?

ADBCE Grid:

A) N<100, which means N can have any value from 1 to 99. Insuff
B) N/20= X+3 --> N= 20X+3 Where N Can have any number of values. INsuff

Using A and B,
N<100 and N = 20X+3

Suff to Find the value of N.

Hence C

Thanks

Yes, Sudhan, but it's not clear from your solution why (1&2) is sufficient. Here's a numerical approach.

Rephrased question:
The possible values of N are 13, 38, 63, 88, 113, 138, 163, 188, 213... what is n?

(1) Not enough information, as there are many values less than 100 in our list.
(2) So, N could be 3, 23, 43, 63, 83, 103, 123, 143, 163, 183, ... Again, we have at least two cases: 63 and 163. Insufficient.
(1&2) This narrows it down to exactly one value: 63. Sufficient.

Another way to deal with it is to express N as functions.

Rephrased question:
N = 25x + 13. What is N? (Here, the variable x can take on integer values. For each x, you'll get a corresponding value of N.)

(1) N < 100 is not sufficient. If x is 1, for example, N is 38 and if x is 3, N is 88.
(2) We are told that N can be expressed as 20y + 3 (where y is an integer). So we can attempt to solve the system:
N = 20y + 3
N = 25x + 13
Can't do it! Only two equations and three variables. Insufficient.

(1&2) Now let's go back to that system from (2). It yields:
20y + 3 = 25x + 13
20y = 25x + 10
So now it's a matter of finding values of x and y that make the equation true. If x = 1, the right side is 35, and that's not a multiple of 20 (as required by the left side). If x = 2, the right side is 60, and that *is* a multiple of 20. So x = 2 and y = 3 is a viable solution. That corresponds to a value of N = 63. Keep going, though. If x = 3, the right side is 85 -- no good, since 88 is not a multiple of 20. If you keep going with this exercise, the next value of x that works is x = 6, since y would equal 8. BUT, x = 6 corresponds to a value of N = 163, which is bigger than 100. Sufficient.

I think you'd agree that for these problems, a numerical solution is probably most efficient. But either way gets you there.

Rey

Rey,

In above solution, once you find first value of N which satisfy both the eqn, You can add LCM of 20 and 25 to first value. So other values of N will be 63, 163, 263 etc
I find this method much faster but want to check whether this can be applied to any remainder equation.

that will work, because you have to add a multiple of N for the remainder upon division by N to stay the same.

since you want both remainders to stay the same, you must add a number that's a multiple of both numbers - so your lcm approach is ideal.

Here it is useful to see that both 20 and 25 are divisors of 100

Multiples of 25 have the tens and units digits 00,25,50,75
So the questions tells us that n ends in either 13,38,63 or 78

Similarly, (2) tells us that n ends in 03, 23, 43,63 or 83

Kevin Armstrong,
Madrid, Spain

Good work everyone.