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mejia401
62. The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A) 1,104
B) 972
C) 672
D) 294
E) 192


Let the middle integer be x
Then the largest integer will be 3x
The smallest integer will be 3x – 23
Sum = x + 3x + 3x – 23 = 40
Then 7x = 63 and x = 9
Hence the numbers are 9(27) (4), since the option must be divisible by 9 hence the answer is (2).
(To check whether a number is divisible by 9, check if the sum of digits is divisible by 9)
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A+B+C = 40
A = C-23
B = C/3
Solving these 3 equations => C = 27, B = 9, A = 4 => ABC = 972
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The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A) 1,104
B) 972
C) 672
D) 294
E) 192

Who can solve this in less than 60 seconds? And how?

The three integers are

a + b + c = 40

3b - 23 + b + 3b = 40

So. 7b = 63

Or, b = 9

So, The three numbers are 4 , 9 & 27

Hence, product of three numbers is 4*9*27 = 972

Thus, the answer will be (B) 972
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3x,x,(3x-23) = three number
3X+X+3X-23=40
X=9
3X=27
3X-23=4


27*9*4=972 (B)
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