Q: What are two consecutive even integers whose sum is 126?

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2 times 30+31+32+33 41+42+43

125, 126 and 127.

127 is between the integers 126 and 128.

The numbers are 41, 42 and 43.

Let the consecutive multiples of 7 be 7x and 7(x+1).7x + 7(x+1) = 1267x + 7x + 7 = 12614x + 7 + 12614x = 126 - 714x = 119x = 119/14x = 8.5For 7x to be a multiple of 7, x must be a whole number, so there are no such consecutive multiples.

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The two even consecutive integers are: 62+64 = 126

There is no solution. Since 126 is even and two consecutive integers have to be one odd and one even number, an odd and an even number cannot summate to an even number.

Divide 126 by 2 and choose the even numbers on both sides.

41+42+43=126.

The numbers are 62 and 64 are two consecutive integers that equal 126.

The integers are are 41, 42 and 43.

41, 42 and 43

41, 42 and 43

The numbers are 62 and 64.

Let the intgers be n- 1 n , & n+1 . This is consecutive. Hence Adding (n-1) + n + ( n +1) = 126 Add the LHS 3n = 126 Divide both sides by '3' n = 42 Hence n- 1 = 41 & n +1 = 43 So the consecutive integers are 41,42, & 43.

There are two consecutive even numbers. The numbers are 62 and 64.

41,42,43