When n is divided by 5 the remainder is 2. When n is divided

Question

When n is divided by 5 the remainder is 2. When n is divided by 7 the remainder is 1. If n is less than 800 , how many values of n are possible?

A. 22 
B. 23 
C. 24 
D. 25 
E. 26

Bunuel · Accepted Answer

First of all it should be mentioned in the stem that n is a positive integer: every GMAT divisibility question will tell you in advance that any unknowns represent positive integers. Given: n=5p+2 , so n can be: 2, 7, 12, 17, 22 , 27, ... n=7q+1 , so n can be: 1, 8, 15, 22 , 29, ... General formula for n based on above two statements will be: n=35k+22 (the divisor should be the least common multiple of above two divisors 5 and 7, so 35 and the remainder should be the first common integer in above two patterns, hence 22). Next, given that 35k+22<800 --> k<22.something --> so k , can take 23 values from 0 to 22, inclusive, which means that n also can take 23 values (the least value of n will be 22 for k=0 ). Answer: B. For more about this concept see: manhattan-remainder-problem-93752.html#p721341 when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html#p722552 when-the-positive-integer-a-is-divided-by-5-and-125591.html#p1028654 Hope it helps.

shaselai · Answer

well we start with 22. Now we need to find the next number that ends in 2 or 7 to satisfy 5's criteria. Since 22 already has remainder of 1 for 7 all we need to do is add 7 to it to find the next number ends in 2 or 7. After finding that number (57) we see the gap is 35. it will keep being 35 since the remainder will keep switching like before until the next N is found. since we know the "gap" is 35 then we multiply by 22( the lowest possible answer) to get 770. Since we know the value starts at 22 and 792 is only a few behind 800 so we stop there. You could also see it as : 22 + 35*x = 800 and find the value that way but I was lazy and used the answer choices (plus it is pretty easy to do 35*22 on the fly). It is probably no coincidence that the gap is 5 X 7 (both values in question are in it) but the number property escapes me so i dont want to feed you the wrong info. But even without it you can get to the 2nd value pretty quick and notice the pattern (can add another 35 to it to double check i guess...).

shaselai · Answer

is it B? I see that the first N is 22 that satisfies the criteria. Then 35 values later 57 satisfies and so on... multiplying 35 X 22 we get 770 and adding the 22 from the beginning we have 23 Ns

You should also notice that N only satisfies 5 if it ends with 2 or 7. Since we know n starts with 22 we can just keep adding 7s to it (to keep 7's remainder) until we hit a number that ends in 2 or 7 to satisfy 5 (7s already satisfied since we kept adding 7s to it). after getting the next one you can then divide it out and find how many "35 number gaps" there are.

gmatjon · Answer

is it B? I see that the first N is 22 that satisfies the criteria. Then 35 values later 57 satisfies and so on... multiplying 35 X 22 we get 770 and adding the 22 from the beginning we have 23 Ns

I see how u get 22, 57 (most likely just by guessing) but I am not able to follow your logic how u find the formula, where u multiply 35 by 22, and get 770 adding 22 u get 23Ns?
Can u explain please in detail?
Thank u

flyingbunny · Answer

for question we can get:
1) n=5x+2
2) n=7y+1

therefore x=(7y-1)/5, so the last digit of 7y is either 1 or 6

when last digit is 1:
y=3, 13, 23...
y=10t+3
then for 2), n=7(10t+3)+1<800
t<11.1, 
so, n could be 12 values.

when last digit is 6:
y=8, 18, 28
y=10k+8
then for 2), n=7(10k+8)+1<800
k<10.6
so n could have 11 values.

in sum, n could have 23 values.

rohansherry · Answer

I dont think you can take it approx 23 ..it has to be 22...... The answer is 23 because it adds the first value ie 22 which the calculation doesnt include.

LenaA · Answer

You can come up with 35 very fast. The difference between the numbers should be LCM (least common multiple) of 5 and 7. Otherwise n will not satisfy both equations : n=5*x+2 and n=7*y+1. As soon as I got 35, I figured out that the first number is 22 (simply by picking numbers). Then (800-22)/35 will be 22.22. So there should be 22 numbers not including the first number and 23 in total.

barakhaiev · Answer

I think the answer is A. 22. 
Here is my solution:

N= 2(mod 5)
N=1(mod 7)

5X+2=N
7Y+1=N

5X+2=7Y+1
5X=7Y-1
X=(7Y-1)/5

min value of Y = 3
min value of N = 22

LCM(7,5) = 35

N=35T+22
35T+4  < (IS LESS)  800
35T<778
T<22.22

Thus, T (number of Ns) is equal to  22,  because the nearest integer, which is less than 778/35 is 22.
However, if we insert the T (22) in a formula N=35T+22, we can see that there are  23  integers on the number line, which is least than 800.

Excuse me for my previous mistake. Definitely  B, 23

MBAhereIcome · Answer

ans: 23 .. for sure

n = 35k+22 .. this is the equation at hand.

35k+22 < 800
k < 22.2
k = 22

total possible numbers = 22+1 = 23 .. we're adding 1 because the value of k starts from 0. here's a list of possible numbers to clear any doubts.

0 -- 22
1 -- 57
2 -- 92
3 -- 127
4 -- 162
5 -- 197
6 -- 232
7 -- 267
8 -- 302
9 -- 337
10 -- 372
11 -- 407
12 -- 442
13 -- 477
14 -- 512
15 -- 547
16 -- 582
17 -- 617
18 -- 652
19 -- 687
20 -- 722
21 -- 757
22 -- 792

keiraria · Answer

hello BRUNUEL 
i understood up to 35K+22 Smaller than 800

BUT how did you get the non inclusive 22 values

thanks so much

best regards

Bunuel · Answer

35k+22<800 --> k<22.something --> so k can be: 0, 1, 2, ..., 21, 22, so it can take total of 23 values. Hope it's clear.

jacket882 · Answer

I don't understand why the remainder should be the first common integer (22).. could you please explain?

Bunuel · Answer

22 is the first number that satisfies both conditions, so in general formula it should be the first possible value of n (for k=0).

gracie · Answer

n=5q+2
n=7p+1
5q+1=7p
least value of q=4
least value of n=22
let x=possible values of n<800
22+(x-1)(5*7)<800
x<23.2
x=23
B

Oppenheimer1945 · Answer

N=5a+2=7b+1
N+13=5(a+3)=7(b+2)
N=35z-13
z=1,2,...23

zmax=813/35=23

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When n is divided by 5 the remainder is 2. When n is divided

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