555

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555

by faraz_jeddah » Sun Sep 22, 2013 2:33 am
Each of the digits 7,5,8,9,4 is used only once to form a 3-digit integer and a 2 digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?

A - One
B - Two
C - Three
D - Four
E - Five
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by faraz_jeddah » Sun Sep 22, 2013 2:38 am
I can only get the pairs as 497 + 58 & 498 + 57
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by vinay1983 » Sun Sep 22, 2013 3:18 am
To get the units digit as 5, among the given digits only 8 and 7 can be used. 7+8 and 8+7, so option B to me.

I think you are right. What is the source and OA?
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by faraz_jeddah » Sun Sep 22, 2013 3:24 am
Nope B is incorrect.

I just solved it.

The other two pairs are 457 + 98 & 458 + 97

OA is D.

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by vinay1983 » Sun Sep 22, 2013 3:52 am
I think today is not my day to get answers correct. Something wrong with me. Will get back refreshed at solving the questions. :(
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by GMATGuruNY » Sun Sep 22, 2013 3:56 am
faraz_jeddah wrote:Each of the digits 7,5,8,9,4 is used only once to form a 3-digit integer and a 2 digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?

A - One
B - Two
C - Three
D - Four
E - Five
Let the 5 digits be A, B, C, D and E, as follows:

ABC
XDE
555

To yield a sum of 555, A=4:

4BC
XDE
555

Given the remaining digits 5, 7, 8, and 9, it is not possible that C+E = 5.
Thus, to yield a units digit of 5, C+E=15.
Implication:
Either C=7 and E=8 or C=8 and E=7.
Total options = 2.

For the tens place, only the digits 5 and 9 remain.
Thus, either B=5 and D=9 or B=9 and D=5.
Total options = 2.

To combine the options above, we multiply:
2*2 = 4.

The correct answer is D.
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by theCodeToGMAT » Sun Sep 22, 2013 4:39 am
For Units place: 7 + 8 = To make sum = _ _ 5

So Numbers:
_ _ 7
x _ 8

For Tens place: need to make sum as or _4_ --> 5 + 9 --> _ 5 5

So, numbers:
_ 5 7
x 9 8

We are only left with 4

So, numbers
4 5 7
x 9 8

However, 5 & 9 and 7 & 8 can swap possitions.. So total possible numbers

4 5 7
x 9 8
4 9 8
x 5 7

Answer[spoiler][D][/spoiler]
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Re: 555

by Scott@TargetTestPrep » Thu Nov 05, 2020 2:05 pm
faraz_jeddah wrote:
Sun Sep 22, 2013 2:33 am
Each of the digits 7,5,8,9,4 is used only once to form a 3-digit integer and a 2 digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?

A - One
B - Two
C - Three
D - Four
E - Five
Solution:

Since the sum of the two integers is 555, the larger (three-digit) integer must be in the 400s.
If the larger integer is 457, then the smaller one is 555 - 457 = 98 (which is formed by the remaining two digits).

If the larger integer is 475, then the smaller one is 555 - 475 = 80 (which is NOT formed by the remaining two digits).

If the larger integer is 458, then the smaller one is 555 - 458 = 97 (which is formed by the remaining two digits).

If the larger integer is 485, then the smaller one is 555 - 485 = 70 (which is NOT formed by the remaining two digits).

If the larger integer is 459, then the smaller one is 555 - 459 = 96 (which is NOT formed by the remaining two digits).

If the larger integer is 495, then the smaller one is 555 - 495 = 60 (which is NOT formed by the remaining two digits).

If the larger integer is 478, then the smaller one is 555 - 478 = 77 (which is NOT formed by the remaining two digits).

If the larger integer is 487, then the smaller one is 555 - 487 = 68 (which is NOT formed by the remaining two digits).

If the larger integer is 479, then the smaller one is 555 - 478 = 76 (which is NOT formed by the remaining two digits).

If the larger integer is 497, then the smaller one is 555 - 497 = 58 (which is formed by the remaining two digits).

If the larger integer is 489, then the smaller one is 555 - 489 = 66 (which is NOT formed by the remaining two digits).

If the larger integer is 498, then the smaller one is 555 - 498 = 57 (which is formed by the remaining two digits).

Answer: D

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