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MathRevolution
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GMAT 1: 760 Q51 V42
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Let the number be ABC

A+B+C = 17
A+C=13
CBA = ABC -99
100C + 10B + A = 100A +10 B +C -99
99A-99C -99 =0
A-C =1
Solving the equation we get
A=7 B=4 C=6

746 E is the answer

Posted from my mobile device
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=>

Assume \(N\) is a three-digit integer represented by \(xyz.\)

Then we have an algebraic expression \(N = 100x + 10y + z.\)

We have \(x + y + z = 17, x + z = 13\) and \(100z + 10y + x = 100x + 10y + z – 99.\)

If we subtract the first \(2\) equations we get \((x + y + z) – (x + z) = 17 – 13\) and \(y = 4.\)

When we substitute this into the last equation, we have \(100z + 10*4 + x = 100x +10*4 + z - 99, 99z – 99x = -99.\) If we multiply by \(-1\) we get \(99x - 99z = 99\) or \(x – z = 1.\)

Then we have \(x = 7\) and \(z = 6,\) since \(x + z = 13\) and \(x – z = 1.\)

So, the original number \(N\) is \(746.\)

Therefore, E is the answer.
Answer: E
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Plugging the answer values seem to be way faster, just reverse the units and 100 th digits and subtract. Will take around 30 secs
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[GMAT math practice question]

\(N\) is a \(3\)-digit positive integer. The sum of all \(3\) digits of \(N\) is \(17\), and the sum of its hundreds digit and its units digit is \(13.\) The new number, which is made by exchanging the digits in the hundreds position and the units digit,\(00000\) is \(99\) less than the original number \(N.\) What is the value of \(N\)?

A. \(944\)

B. \(449\)

C. \(548\)

D. \(845\)

E. \(746\)

First thing, can someone confirm what are those 000000?

The options give you the reverse so you dont even have to take effort.
Clearly between A and B & between C and D the difference is greater than 99.

That leaves us with E.

53 seconds :)
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