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Q: Find 4 consecutive even integers where the product of the two smaller numbers is 72 less than the product of the larger two integers?

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-1

The numbers are 5, 7, 9, 11 (35 is 64 less than 99) The equation is (A)(A+2) = (A+4)(A+6) - 64 A2 + 2A = A2 +10A +24 -64 2A = 10A - 40 8A = 40 A = 5

x+3 and x+4 would be consecutive integers.

5, 6 and 7

12, 13 and 14

Related questions

4,6,8,10

The numbers are 11, 13, 15 and 17.

The product of 2 consecutive positive number is 48. Find the 2 numbers

The smaller of the two numbers is 31.

13 and 12 are the two integers that have the product of 156 and 12 is the smaller of the two.

Smaller number is '6'

It is 23.

You can solve this in two ways.1) Trial and error. That is, try multiplying two consecutive integers; if the product is too large, try smaller integers; if the product is too small, try larger consecutive integers. 2) Call the two consecutive integers "n" and "n+1", and solve the equation: n(n+1)=210

44 & 45

-1

The numbers are 9 and 10.

6 x 8 = 48 10 x 12 = 120 120 - 48 = 72

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