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Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

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Let three consecutive integers be x, x + 1, x + 2. According to the question, 2x + 3(x + 1) + 4(x + 2) = 74 2x + 3x + 3 + 4x + 8 = 74 9x + 11 = 74 On transposing 11 to R.H.S, we obtain 9x = 74 − 11 9x = 63 On dividing both sides by 9, we obtain x = 7 x + 1 = 7 + 1 = 8 x + 2 = 7 + 2 = 9 Hence,...
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Let three consecutive integers be x, x + 1, x + 2. According to the question,

2x + 3(x + 1) + 4(x + 2) = 74

2x + 3x + 3 + 4x + 8 = 74

9x + 11 = 74

On transposing 11 to R.H.S, we obtain

9x = 74 − 11

9x = 63

On dividing both sides by 9, we obtain

x = 7

x + 1 = 7 + 1 = 8

x + 2 = 7 + 2 = 9

Hence, the numbers are 7, 8, and 9.

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Engineering graduate from NIT Calicut, Rank holder (#47) in State Entrance Examination.

Let x be the first integer. second and third integers are x+1,x+2 respectively. 2x+3(x+1)+4(x+2)=74 2x+3x+3+4x+8=74 9x+11=74 9x=74-11=63 x=63/9=7 numbers are 7,8,9
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