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Re: remainders [#permalink]
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yezz wrote:
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

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.
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Re: remainders [#permalink]
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
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Re: remainders [#permalink]
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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.
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Re: remainders [#permalink]
GMATBLACKBELT wrote:
shaselai wrote:
yezz wrote:
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

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.


I came up with 35 as well after writing it out. No way I could ever do this in 2min.

Interestingly after coming up with 35... and plowing through to the answer I get 23. An easier way... that may work (not sure whether its full proof). 800/35= 22.857=~23...


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.
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Re: remainders [#permalink]
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GMATBLACKBELT wrote:
shaselai wrote:
yezz wrote:
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

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.


I came up with 35 as well after writing it out. No way I could ever do this in 2min.

Interestingly after coming up with 35... and plowing through to the answer I get 23. An easier way... that may work (not sure whether its full proof). 800/35= 22.857=~23...




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.
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Re: remainders [#permalink]
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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

Originally posted by barakhaiev on 02 Sep 2009, 10:14.
Last edited by barakhaiev on 02 Sep 2009, 12:51, edited 1 time in total.
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Re: remainders [#permalink]
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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
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Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
Bunuel wrote:
yezz wrote:
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


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.


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
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Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
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keiraria wrote:
Bunuel wrote:
yezz wrote:
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


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.


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


\(35k+22<800\) --> \(k<22.something\) --> so \(k\) can be: 0, 1, 2, ..., 21, 22, so it can take total of 23 values.

Hope it's clear.
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Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
Quote:
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).


I don't understand why the remainder should be the first common integer (22).. could you please explain?
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Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
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jacket882 wrote:
Quote:
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).


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


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).
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When n is divided by 5 the remainder is 2. When n is divided [#permalink]
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yezz wrote:
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


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

Originally posted by gracie on 03 Jun 2017, 22:59.
Last edited by gracie on 08 Dec 2017, 17:54, edited 1 time in total.
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Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
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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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Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
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