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3 digit integer and 2 digit integer

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3 digit integer and 2 digit integer

by yass20015 » Fri Aug 21, 2015 12:25 am
I get this one in mba.com, but I solve it in 4+ min... Do anyone has a quicker method ?
Each of the digits 7,5, 8, 9 and 4 is used only once to form a three-digit integer and a two-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 GMATGuruNY » Fri Aug 21, 2015 4:27 am
yass20015 wrote:Each of the digits 7,5, 8, 9 and 4 is used only once to form a three-digit integer and a two-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 Max@Math Revolution » Fri Aug 21, 2015 8:26 pm
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


I get this one in mba.com, but I solve it in 4+ min... Do anyone has a quicker method ?
Each of the digits 7,5, 8, 9 and 4 is used only once to form a three-digit integer and a two-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

==> each digits can only use 7,5,8,9,4 once, thus abc(3digit integer)+de(2digit integer)=555 leads to a=4, b+d=14, c+e=15. Then, (a,b,c,d,e)=(4,5,7,9,8),(4,5,8,9,7),(4,9,7,5,8),(4,9,8,5,7) thus there are 4 forms. Therefore D is the answer.


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by Matt@VeritasPrep » Sun Aug 23, 2015 11:13 am
We know that the three digit integer must begin with 4 or 5. Let's try each option.

Case I:: The three-digit integer begins with 5

This seems reasonable, but the smallest three-digit we can make is 547, so there are no such solutions.

Case II:: The three-digit integer begins with 4

This is fine. Now let's look at our units digits. The two units digits must sum to a number that ends in 5, since 555 ends in 5. So we could only have (8 + 7). This means we have

4a7 + b8 = 555
or
4a8 + b7 = 555

The only digits left for a and b are 5 and 9, which work in both equations in either place. (9 + 5 + 1 -- the 1 we carry from 8+7 -- is 15, so we'll always get what we want.)

This gives us all the solutions:

457 + 98
497 + 58
458 + 97
498 + 57
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