When an integer divided by 35, remainder is 2, when divided

Question

When an integer divided by 35, remainder is 2, when divided by 31, remainder is 3. What is the number? A list of answers are given
Answer: 282

Tried the chinese remainder on this one worked but took too long is there any other way to solve this problem that will save time?

kevincan · Answer

I don't know what the Chinese remainder theorem is, but this method isn't too slow:

We know our sought after number x can be written as 35m+2 or 31k+3, where k and m are both integers. 

So 35m+2=31k+3

35m-1=31k

k= (35m-1)/31= (4m-1)/31 +m

Since k and m are an integers , 4m-1 must be a multiple of 31

m=8 is the smallest positive value that will do the trick- k would be 9

35(8)+2=282=31(9)+3

kevincan · Answer

By the way, there are more numbers that would yield the required remainders. I would not do this in exam, though. I would divide all the choices that end in 2 or 7 by 31 (to see which yield a remainder of 3), and then test the choices that passed this test by 35 (to see which yield a remainder of 2)

apollo168 · Answer

Hi Kevin

From this pt (35m-1)/31 how did u arrive at this (4m-1)/31 +m?

apollo168 · Answer

Sorry tried again got it. Haven't seen anyone use that method in a long time.

lan583 · Answer

I don't know what the Chinese remainder theorem is, but this method isn't too slow: 

How did you arrived at the computation below. I am talking about the numbers in red. Thanks. 

We know our sought after number x can be written as 35m+2 or 31k+3, where k and m are both integers. 

So 35m+2=31k+3 

35m-1=31k 

k= (35m-1)/31=   (4m-1)/31 +m   

Since k and m are an integers , 4m-1 must be a multiple of 31 

m=8 is the smallest positive value that will do the trick- k would be 9 

35(8)+2=282=31(9)+3

kevincan · Answer

(35m-1)/31= (31m+4m-1)/31

skarya · Answer

When an integer divided by 35, remainder is 2, when divided by 31, remainder is 3. What is the number? 

x=35a + 2 => a=(x-2)/35 --> (1)
x=31b + 3 => b=(x-3)/31 --> (2)

where a and b are integers

(1) => (x-2) is divisible by 35 i.e. divisible by 5 and 7 
hence last digit must be 5 or 0 and it should also be divisible by 7
=>x-2 = 35, 70, 105, 140, ....
=> x=37, 72, 107, 142,... (last digit of x is 7 or 2)

(2) => (x-3) is divisible by 31 (a prime number) 
x-3= 124, 279, ... => x= 127, 282...

x= 282 satisfy both conditions.

prashrash · Answer

why only numbers ending in 2 and 7 please ?

When an integer divided by 35, remainder is 2, when divided

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards