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Q: Find 4 consecutive even integers where the product of the smaller two numbers is 56 less than the product of the two larger numbers?

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Related questions

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

The numbers are 11, 13, 15 and 17.

They are 6, 8, 10 and 12.

The smaller of the two numbers is 31.

It is 23.

Smaller number is '6'

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

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

Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.

the answer is 6

In 'normal' arithmetic, there is no solution of 3 consecutive odd numbers where the product of the smaller two is 22 less than that of the larger two. For instance difference in products for 1-3-5 is 12, for 3-5-7 it is 20, and for 5-7-9 it is 28. The series steps by 8 integers for each set of 3 odd numbers investigated.

The numbers are 9 and 10.

48 = 6 x 8, so the smaller is 6.

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

Divide the sum of the two consecutive even integers by 2: 90/2 = 45. The smaller of these integers will be one less than 45 and the larger will be one more than 45, so the two consecutive even integers will be 44 and 46.

6,8,10,12

The four even integers are 6, 8, 10, 12. 6 x 8 = 48 10 x 12 = 120 120 - 48 = 72

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

There are no two consecutive even integers, consecutive odd integers, or consecutive integers that satisfy that relationship.

5,6,7

11,12,13

This is best solved by trial-and-error. If one set of consecutive even integers doesn't work, try a different set. Hint: The integers involved are fairly small.

8,10,12