Question

Jenna writes four three-digit numbers:
634 995 816 742
Later, for each of the four numbers, she multiplies the $2^{nd}$ digit with the $3^{rd}$ digit and adds the product with the square of the $1^{st}$ digit.

Find which number will be the last number if the numbers are then arranged in ascending order based on the sums.

• A. 634
• B. 816
• C. 995
• D. 742

Answer 1

The correct option is C 995

Let's analyze the digits one by one in the following way:

634:
Multiplying the $2^{nd}$ digit with the $3^{rd}$ digit:
$3\times 4=12$
Square of $1^{st}$ digit = 36
∴ Sum = 12 + 36 = 48

995:
Multiplying the $2^{nd}$ digit with the $3^{rd}$ digit:
$9\times 5=45$
Square of $1^{st}$ digit = 81
∴ Sum = 45 + 81 = 126

816:
Multiplying the $2^{nd}$ digit with the $3^{rd}$ digit:
$1\times 6=6$
Square of $1^{st}$ digit = 64
∴ Sum = 6 + 64 = 70

742:
Multiplying the $2^{nd}$ digit with the $3^{rd}$ digit:
$4\times 2=8$
Square of $1^{st}$ digit = 49
∴ Sum = 8 + 49 = 57

Arranging the sums in ascending order, we get:

48 57 70 126

So, arranging the numbers corresponding to the sums, we get:
634 742 816 995

Thus, 995 will be the last number in the arrangement.