When a 2-digit positive integer N is divided by 2, the remainder is 1

Question

When a 2-digit positive integer N is divided by 2, the remainder is 1, when it is divided by 3 the remainder is 2, and when it is divided by 5 the remainder is 4. What is the sum of all possible values of N?

A. 155
B. 166
C. 177
D. 188
E. 200

A. 155
B. 166
C. 177
D. 188
E. 200

Crater · Accepted Answer

When a 2-digit positive integer N is divided by 2, the remainder is 1, when it is divided by 3 the remainder is 2, and when it is divided by 5 the remainder is 4. What is the sum of all possible values of N?

A. 155
B. 166
C. 177
D. 188
E. 200

The concept is very simple: If difference between divisor and remainder is constant, take LCM and subtract 1 from it.
eg, 
Divisor - 2,3,5
Remainder - 1,2,4
2-1=3-2=5-4

In this case, LCM of 2,3,5 = 30
Common difference = 1
Total two digit numbers possible, 29, 59, 89 
Sum=177

Option (C)

Can,Will · Answer

Its C. 
LCM of 2,3,and 5 is 30 .
desired numbers are 29,59 and 89
sum 177

Archit3110 · Answer

given
n=2a+1
n=3b+2
n=5c+4
n has to be unit digit 9 
possible n values ; 29, 59, 89 sum is 177
option C

jessicatherrien · Answer

I'm confused why the common difference is one.

According to this post: https://gmatclub.com/forum/a-person-inh ... 32500.html

when we have multiple equations, we take the LCM and the lowest integer --- 30 (LCM)q+lowest integer is (???)

How would we use this approach to find the common difference is one? Thanks!

laborumpossimus · Answer

i am not sure if my approach is correct but i evaluated all the options and ABDE were divisible by either 2 , 3 or 5 so i eliminated them

egmat · Answer

Hi laborumpossimus,

I can see exactly what you did, so let me point out the one gap before you rely on this again.

Your instinct about  individual N values  is fine: since N leaves a nonzero remainder when divided by  2 ,  3 , and  5 , each valid N ( 29, 59, 89 ) really is  not  divisible by  2 ,  3 , or  5 .

But here's the catch:  the answer choices are not N - they are the  sum  of all the N values. And a sum of numbers that aren't divisible by  3  can absolutely be divisible by  3 .

Check the correct answer itself:  177 = 3 × 59 . It  is  divisible by  3 . So by your own rule - "eliminate anything divisible by  2 ,  3 , or  5 " - you would have thrown out  C  too. The fact that only C survived your elimination was a coincidence, not a valid deduction.

The safe path is the one already shown in the thread: each remainder is exactly  one less  than its divisor, so  N + 1  is a multiple of lcm( 2 , 3 , 5 ) =  30 . That gives N = 30k - 1, i.e.  29, 59, 89 , and their sum is  177 .

Quick way to feel the flaw: 
- Take  1  and  2  - neither is divisible by  3 .
- Their sum is  3 , which  is  divisible by  3 .

So divisibility of the parts tells you nothing guaranteed about divisibility of the sum. Use that elimination idea only on N itself, never on a total built from several N's.

Answer: C

laborumpossimus · Answer

Ohh yes that nakes sense i am gonna use the LCM approach 
Also what about question where in you dont have common difference 
How can we approach them in the easiest possible way

When a 2-digit positive integer N is divided by 2, the remainder is 1

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When a 2-digit positive integer N is divided by 2, the remainder is 1

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