A Product Manipulation

Algebra Level 2

Suppose that a, b, c, d are positive integers. There are 6 possible products obtained by pairing them (namely ab, ac, ad, bc, bd, and cd). The values of 5 products, in a random order, are 2, 3, 5, 6, and 10. What could be the sixth product?

The answer is 15.

2 solutions

Tarikul Islam
May 30, 2020

In this problem, a is present in 3 products, and the same goes for b, c, and d.

2= 1 times 2; 3= 1 times 3; 5= 1 times 5; 6= 2 times 3; 10= 2*5

So, now, if we take 15= 3*5, then 1, 2, 3, 5 each will be present in three products. So, since we are working on the positive integers, the sixth product must be equal to 15.

You stated $b$ , $c$ , and $d$ are present in 3 products. If we consider $ab, ac, ad, bc, bd$ , we see that there are $2$ $c$ s and $2$ $d$ s in these products. Also, please clarify what you did in the second sentence.

Atomsky Jahid - 1 year ago

I have edited my second line, and still do not know how to write with Latex. Now, do you have any questions?

Tarikul Islam - 1 year ago

Atomsky Jahid
May 30, 2020

There are three ways to get $abcd$ , namely $ab$ multiplied by $cd$ , $ac$ multiplied by $bd$ and $ad$ multiplied by $bc$ . Hence, we can find two pairs which will give us $abcd$ . $3 \times 10 = 5 \times 6 = 30$ Now, let's assume the sixth product is $x$ . Then, $2x=3 \times 10 = 5 \times 6 =30 \\ \implies x= \boxed{15}$

A Product Manipulation

The answer is 15.

2 solutions

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