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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Question Stats: 28% (02:54) correct 72% (02:09) wrong based on 111 sessions

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When integer A is divided by 79, remainder is X and when integer B is divided by 79, remainder is Y. Given that X + Y < 79. What is X * Y?

1) When A + B is divided by 79, remainder is 18
2) When A * B is divided by 79, remainder is 77

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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Harley1980 wrote:
When integer A is divided by 79, remainder is X and when integer B is divided by 79, remainder is Y. What is X * Y?

1) When A + B is divided by 79, remainder is 18
2) When A * B is divided by 79, remainder is 77

Second is not sufficient. X*Y could be 77 or 156 for example.
A=86 X=7
B=90 Y=11

then A*B=7740=79*97+77, and X*Y=77

A=83 X=4
B=118 Y=39

then A*B=9794=79*123+77, but X*Y=156.

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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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2
A = 79n+X
B = 79m+Y

1. A+B = 79(n+M) +X+Y. After division by 79 we have X+Y = 18.
2. A*B = (79n+X)*(79m+Y). Only X*Y is not divided by 79. So X*Y = 77

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GMAT 1: 740 Q50 V40 Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Great question!

Perfect for people who are having trouble with remainder formulas.
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Harley1980 wrote:
When integer A is divided by 79, remainder is X and when integer B is divided by 79, remainder is Y. What is X * Y?

1) When A + B is divided by 79, remainder is 18
2) When A * B is divided by 79, remainder is 77

Hi,

I used simpler numbers (11, 19, 4, 8 and divisor as 3) to work up examples and arrived at the correct answer.

Could you please help me with the properties Thanks,
aimtoteach
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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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1
aimtoteach wrote:
Harley1980 wrote:
When integer A is divided by 79, remainder is X and when integer B is divided by 79, remainder is Y. What is X * Y?

1) When A + B is divided by 79, remainder is 18
2) When A * B is divided by 79, remainder is 77

Hi,

I used simpler numbers (11, 19, 4, 8 and divisor as 3) to work up examples and arrived at the correct answer.

Could you please help me with the properties Thanks,
aimtoteach

Hello aimtoteach.

When we have task about division different numbers on the same divisor we can sum or multiple the equations. For example
"When 7 divided by 5 , remainder is 2 and when 11 divided by 5 , remainder is 1" We can write this information in such way:
$$7=5∗X+2$$
$$11=5∗Y+1$$

if we sum this equations we received such result
$$7+11=5∗Z+(2+1)$$ so $$18 / 5$$ give us remainder 3

of if we multiple this equations we received:
$$7∗11=5∗R+2$$ so $$77 / 5$$ give us remainder 2

In our task we can write this phrase "When integer A is divided by 79, remainder is X and when integer B is divided by 79, remainder is Y" in such way:
$$A=79s+X$$
$$B=79r+Y$$

and when we sum this equations we received
$$A+B=79z+(X+Y)$$
and from first statement we know that "When A + B is divided by 79, remainder is 18"
So we can make infer that $$X+Y=18$$. But from this inforamation we can't make infer about what X*Y equal because X+Y = 18 can give us more than one combinations: 1+17; 2+16 etc.
Insufficient

and when we multiple this equations we received
$$A∗B=79z+(X∗Y)$$
and from second statement we know that "When A * B is divided by 79, remainder is 77"

So we can make infer that X∗Y=77 and we find answer that we need
Sufficient.

Update (thanks to smyarga for noticing mistake in answer):
This is insufficient statement. Because when we add or multiply such equations, we should take into account that X+Y or X*Y can give bigger result than divisor. And in such cases we should subtract divisor from this result and only after this it will give us correct remainder.
In our task there is possible such variant:
$$A=83$$ and $$X=4$$; when $$83$$ divided by $$79$$ there is remainder $$4$$
$$B=118$$ and $$Y=39$$; when $$118$$ diveded by $$79$$ there is remainder $$39$$

and when we multiple $$A$$ and $$B$$ we received $$9794$$ and divided by $$79$$ we will have remainder $$77$$ but $$X*Y=156$$ and this product not equal to remainder
So information in second statement is not sufficient and we need first statement too.

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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Hi Harley 1980,

Harley1980 wrote:
When integer A is divided by 79, remainder is X and when integer B is divided by 79, remainder is Y. What is X * Y?

1) When A + B is divided by 79, remainder is 18
2) When A * B is divided by 79, remainder is 77

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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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shriramvelamuri wrote:
Hi Harley 1980,

Harley1980 wrote:
When integer A is divided by 79, remainder is X and when integer B is divided by 79, remainder is Y. What is X * Y?

1) When A + B is divided by 79, remainder is 18
2) When A * B is divided by 79, remainder is 77

Hello shriramvelamuri

Statement 2 is insufficient because there is possible at least two variants:
$$A=86$$ and $$X=7$$; when $$86$$ divided by $$79$$ there is remainder $$7$$
$$B=90$$ and $$Y=11$$; when $$90$$ divided by $$79$$ there is remainder $$11$$
and when we multiple $$A$$ and $$B$$ we received $$7740$$ and divided by $$79$$ we will have remainder $$77$$ and $$X*Y = 77$$

But we can have another scenario:
$$A=83$$ and $$X=4$$; when $$83$$ divided by $$79$$ there is remainder $$4$$
$$B=118$$ and $$Y=39$$; when $$118$$ diveded by $$79$$ there is remainder $$39$$

and when we multiple $$A$$ and $$B$$ we received $$9794$$ and divided by $$79$$ we will have remainder $$77$$ but $$X*Y=156$$ and this product not equal to remainder
So information in second statement is not sufficient and we need first statement too.

Thanks to smyarga she is clarify situation and give correct answer and explanation to this task.
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Heey Harley,

I think that the first statement is also incorrect, if we consider A= 149 (A=(1x79) + 70) and B = 106 (1x79 + 27); in this case (and in an infinity of other similar cases) A+B = 255 = (3x79) + 18, BUT X+Y = 97 and not 18 !! So it makes no sense to infer that X+Y = 18, thus the correct answer for this question is E.
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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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issamL wrote:
Heey Harley,

I think that the first statement is also incorrect, if we consider A= 149 (A=(1x79) + 70) and B = 106 (1x79 + 27); in this case (and in an infinity of other similar cases) A+B = 255 = (3x79) + 18, BUT X+Y = 97 and not 18 !! So it makes no sense to infer that X+Y = 18, thus the correct answer for this question is E.

Hello issamL

We do not infer that X + Y = 18. This is fact from the first statement. So you just take wrong parameteres that contradict to first statement.
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Thank you harley,

but let me prove why it's not a fact.

A= 79 k + X;
B= 79 z + Y;
A + B = 79(k+z) + (X+Y)
;
Statement 1 says " When A + B is divided by 79, remainder is 18" : A+B = 79m + 18;

by stating X+Y = 18, we assume that (k+z) = m and that's not always true, so it's not a fact.
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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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issamL wrote:
Thank you harley,

but let me prove why it's not a fact.

A= 79 k + X;
B= 79 z + Y;
A + B = 79(k+z) + (X+Y)
;
Statement 1 says " When A + B is divided by 79, remainder is 18" : A+B = 79m + 18;

by stating X+Y = 18, we assume that (k+z) = m and that's not always true, so it's not a fact.

Sorry issamL

I answer in hurry and miss you point.

Yes we can choose such numbers that will give result 18 and another numbers that will give us result 79 + 18.
And because of this 1 statement is insufficient.

But why you decide that both statement are too insufficient and answer is E?
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Because there are more variables than equations;

X+Y = 79k + 18;
X*Y= 79z + 77

with both k and z belonging to {0,1,........N}.
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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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issamL wrote:
Because there are more variables than equations;

X+Y = 79k + 18;
X*Y= 79z + 77

with both k and z belonging to {0,1,........N}.

Hello issamL
Looks like you are right. Thank you.
Weird that nobody pay attention on this early.
I changed the task (add part "Given that X + Y < 79"), could you please try this new variant?
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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issamL wrote:
Because there are more variables than equations;

X+Y = 79k + 18;
X*Y= 79z + 77

with both k and z belonging to {0,1,........N}.

Since X and Y are remainders when divided by 79, they must each be less than 79. X<79 and Y<79, therefore X+Y<158 The only values that X+Y can take in this problem is 18 or 97. For all the integer values X from X=0 to X=18 and Y from Y=0 to Y=18 that satisfy the sum equal to 18, only X=7 and Y=11 or X=11 and Y=7 satisfy X+Y = 79k + 18 and X*Y= 79z + 77 There is no such solution for sum equal 97 that satisfies X+Y = 79k + 18 and X*Y= 79z + 77 without violating X<79 and Y<79. I wrote a simple program to find that out.
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GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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issamL wrote:
Because there are more variables than equations;

X+Y = 79k + 18;
X*Y= 79z + 77

with both k and z belonging to {0,1,........N}.

Since X and Y are remainders when divided by 79, they must each be less than 79. X<79 and Y<79, therefore X+Y<158 The only values that X+Y can take in this problem is 18 or 97. For all the integer values X from X=0 to X=18 and Y from Y=0 to Y=18 that satisfy the sum equal to 18, only X=7 and Y=11 or X=11 and Y=7 satisfy X+Y = 79k + 18 and X*Y= 79z + 77 There is no such solution for sum equal 97 that satisfies X+Y = 79k + 18 and X*Y= 79z + 77 without violating X<79 and Y<79. I wrote a simple program to find that out.

But what about case when X = 70 and Y = 27?
X + Y = 97 = 79 + 18
X * Y = 1890 = 79 * 23 + 73
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Harley1980 wrote:
issamL wrote:
Because there are more variables than equations;

X+Y = 79k + 18;
X*Y= 79z + 77

with both k and z belonging to {0,1,........N}.

Hello issamL
Looks like you are right. Thank you.
Weird that nobody pay attention on this early.
I changed the task (add part "Given that X + Y < 79"), could you please try this new variant?

Hey Harley,

You are welcome.. It's definitely better with the new constraint X+Y < 79 as the only 2 pairs of X & Y that we can have are (7,11) and (11,7), therefore the only possible value of X*Y is 77. so answer (C).
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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Harley1980 wrote:
issamL wrote:
Because there are more variables than equations;

X+Y = 79k + 18;
X*Y= 79z + 77

with both k and z belonging to {0,1,........N}.

Since X and Y are remainders when divided by 79, they must each be less than 79. X<79 and Y<79, therefore X+Y<158 The only values that X+Y can take in this problem is 18 or 97. For all the integer values X from X=0 to X=18 and Y from Y=0 to Y=18 that satisfy the sum equal to 18, only X=7 and Y=11 or X=11 and Y=7 satisfy X+Y = 79k + 18 and X*Y= 79z + 77 There is no such solution for sum equal 97 that satisfies X+Y = 79k + 18 and X*Y= 79z + 77 without violating X<79 and Y<79. I wrote a simple program to find that out.

But what about case when X = 70 and Y = 27?
X + Y = 97 = 79 + 18
X * Y = 1890 = 79 * 23 + 73

Your example is not valid. The remainder is supposed to be 77, not 73. The answer is C regardless of your modification. I will post the program as soon as I can. I need to have 5 posts before I can post a hyperlink.
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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issamL wrote:
Harley1980 wrote:
issamL wrote:
Because there are more variables than equations;

X+Y = 79k + 18;
X*Y= 79z + 77

with both k and z belonging to {0,1,........N}.

Looks like you are right. Thank you.
Weird that nobody pay attention on this early.
I changed the task (add part "Given that X + Y < 79"), could you please try this new variant?

Hey Harley,

You are welcome.. It's definitely better with the new constraint X+Y < 79 as the only 2 pairs of X & Y that we can have are (7,11) and (11,7), therefore the only possible value of X*Y is 77. so answer (C).

I agree that the new constraint, X+Y<79, makes the problem better. I think before the new constraint, the problem was out of GMAT scope because it would be too time consuming to figure out whether there exists two numbers that add up to 97 whose product is a number that is 77 more than a multiple of 79. I had to write my own simple program to find out that there is no such numbers.
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Re: When integer A divide by 79 remainder is X and when integer B divided  [#permalink]

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issamL wrote:
Because there are more variables than equations;

X+Y = 79k + 18;
X*Y= 79z + 77

with both k and z belonging to {0,1,........N}.

Since X and Y are remainders when divided by 79, they must each be less than 79. X<79 and Y<79, therefore X+Y<158 The only values that X+Y can take in this problem is 18 or 97. For all the integer values X from X=0 to X=18 and Y from Y=0 to Y=18 that satisfy the sum equal to 18, only X=7 and Y=11 or X=11 and Y=7 satisfy X+Y = 79k + 18 and X*Y= 79z + 77 There is no such solution for sum equal 97 that satisfies X+Y = 79k + 18 and X*Y= 79z + 77 without violating X<79 and Y<79. I wrote a simple program to find that out.

Here is the program: http://www.codeskulptor.org/#user40_woK7tiG7iVeMdz0.py
The sum should be changed to 18 or 97. Re: When integer A divide by 79 remainder is X and when integer B divided   [#permalink] 25 Aug 2015, 17:02

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