Another Problem

[yoshiki]

The product of two whole numbers is 10,000,000. None of them is a
multiple of ten. What is the sum of both the numbers?
A – 20, B – 78253, C – 4264, D – 29574, E – 35233

[VBAhack]

Just try all of the answers to see which one is correct. Here is the methodology:

xy = 1e7
x+y = c
x = c-y
(c-y)y = 1e7
cy - y² = 1e7
y² - cy + 1e7 = 0

Use quadratic equation to solve for y. Try the given values of c to see which one works (B – 78253). .

[Logophobic]

No, that's not it. B is correct, but the solution is much simpler.

10,000,000 = 10⁷ = 2⁷ * 5⁷

Since neither of the numbers in question is a multiple of 10, the only solution is 128 + 78125 = 78253

[NotLKH]

since 10,000,000 is = (2*5)⁷

Then if the 2 unknown numbers are represented as A and B, such that

A = 2^x*5^y
B = 2^(7-x)*5^(7-y)

Then,
If A and/or B was divisible by 10 then x and y would both have to at least 1.

But since you indicate that A and B are both not divisible by 10, then
either x = 0 and y = 7, or verse visa.

Either way,

the 2 numbers that multiply to 10,000,000 where neither are divisible by 10 are 2⁷ and 5⁷

and their sum is 78253.


[EDIT] LogoPhobic! [/EDIT]

[VBAhack]

No, that's not it. B is correct, but the solution is much simpler.

Nice, elegant solution. But I don't understand your comment. What isn't it?

[Logophobic]

If the solution was not given as multiple-choice, your method would be of no use:

The product of two whole numbers is 583,443. Neither number is a
multiple of 21. What is the sum of these two numbers?

[opus]

If the solution was not given as multiple-choice, your method would be of no use:

In which deductive way did you come to your solution?

[VBAhack]

If the solution was not given as multiple-choice, your method would be of no use

True, but since the answers were given, it seems like a perfectly valid approach to me. Just using the information provided.....