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a, b, c are prime numbers that satisfy: a > b > c a>b>c a + b + c = 52 a+b+c=52 a b c = 22 a-b-c=22 What is the value of a b c abc ?


The answer is 962.

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5 solutions

Morgan Blake
Jan 1, 2014

Summing the two equations: 2 a = 73 2a=73 a = 37 a=37

Therefore, b + c = 15 b+c=15 .

The sum of an odd number and an even number is odd: one of b and c must be odd and the other even.

There is only one even prime: 2 (which is also the smallest). Therefore, c = 2 c=2 .

From that, b = 13 b=13 .

Finally, a b c = 37 × 13 × 2 = 962 abc = 37 \times 13 \times 2 = \boxed{962}

2a=74

sapan karodia - 7 years, 5 months ago

Sorry, typo!

Morgan Blake - 7 years, 5 months ago
Krishna Upman
Jan 6, 2014

37+13+2=52 37-13-2=22 now 37 13 2=962

Budi Utomo
Jan 1, 2014

(a+b+c) - (a-b-c) = 52-22 ---> 2b + 2c = 30 --> 2(b+c) = 30 ----> b+c = 15 ----> (a+b+c) - (b+c) = 52 - 15 ---> a= 37. Because b and c are prime numbers so 15 = 13 + 2 ---> b>c so b= 13 and c = 2 . Thus abc = 37.13.2 = 962

Rahul Patel
Jan 1, 2014

a+(b+c)=52 a-(b+c)=22 a=37 b+c=15 hence b=13 and c=2

Milly Choochoo
Jan 1, 2014

You can shorten your prime number search by realizing that a ( 52 a ) = 22 a- ( 52 - a ) = 22 . By starting at 47 47 and working down the primes, you quickly realize that a = 37 a = 37 , and thus b = 13 b = 13 and c = 2 c = 2 .

37 × 13 × 2 = 962 37 \times 13 \times 2 = 962

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