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Q: Find three consecutive integers with sum 402?

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402, 403, 404

Yes: 398, 399, 400, 401 and 402.

2 x 3 x 67 = 402 2 x 201 = 402 3 x 134 = 402 6 x 67 = 402 To solve such a problem you use prime decomposition: This allows you to break any integer down into a multiple of prime numbers as in 2x3x67=402. Two, Three and 67 are all prime. From there to find all possible options, multiply the primes together: 2x3x67=402 --> 3x67=201 --> 2x201=402 2x3x67=402 --> 2x67=134 --> 3x134=402 2x3x67=402 --> 2x3=6 --> 6x67=402 This is under the assumption that this problem is restricted to integers. If real numbers are used, there are infinitely many solutions:

398, 399, 400, 401 and 402 is one possible set.

Three sig(nificant) fig(ure)s in '402'.

It is 260.

2010

on page 402

402 grams = .402 kg

402/1000

positive 402. the absolute value makes any number positive, even if it already is. So the absolute value of -402 is 402. the absolute value of 402 is also 402.

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402-197 = 205

402

402 of them.

402 201,2 67,3,2

402 = CDII

402

There are 40.2 tens in 402.

402/1

402

The factors of 402 are: 1, 2, 3, 6, 67, 134, 201, 402.

67

Answer: 402 m = 1318.897 '

402 = 2 * 3 * 67