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Q: What two positive integers have a sum of 141 and a product of 4914?

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The product of 141 multiplied by 47 is 6627

two prime numbers whose product is 141 = 3 & 47

The numbers are 45, 47, and 49.

The prime factors are 3 and 47: So, 3 x 47 = 141

The greatest common multiple of any set of integers is infinite.

As 1+4+1=6 then 141 can be divided by 3 141=3 x 47 (47 is a prime number)

I don't think there's a solution to this problem. Let's work it through.Two consecutive integers: x & y. So y = x + 1.Product is 34 more than sum. Product (x*y) = sum(x+y) +34Substitute [y=x+1] into second equation: x*(x+1) = x + (x+1) + 34x2 + x = x + x + 1 + 34 --> x2 + x - 2*x - 35 = 0 --> x2 - x - 35 = 0 (quadratic)In the quadratic formula: a = 1, b = -1, c = -35, so we have (1 Â± sqrt[1 - 4*(-35)])/(2*1) = (1 Â± sqrt(141))/2; sqrt(141) is irrational, so the solution is not integers. Someone else can maybe check my work.

141

Possible the simplest math question ever. -420 / 3 = -140. So the numbers you need are -139 , -140 , -141.

This would be impossible. The closest result you can get is 45 + 47 + 49 = 141

All integers greater than 1 are one or the other.

141

1, 3, 5, 15, 47, 141, 235, 705.

141 x 141 = 19,881

The prime factors of 1,269 are: 3 and 47. (3 x 47 = 141)

5% of 141= 5% * 141= 0.05 * 141= 7.05

16% of 141 = 16% * 141 = 0.16 * 141 = 22.56

The positive integer factors of 987 are: 1, 3, 7, 21, 47, 141, 329, 987

The positive integer factors of 1269 are: 1, 3, 9, 27, 47, 141, 423, 1269

sum means numbers which add up to 141 41 is an obvious one which leaves you with 100 simplest thing would to be divide 100 by 2 but that gives 50 which is an even number so use 49 and 51. 41 + 49 + 51 = 141 Note, these aren't the only 3 odd numbers which sum to 141, that just seemed the easiest way to work it out to me.

15% of 141 = 21.15= 15%/100% * 141= 0.15 * 141= 21.15

141 thousandths is 141 × 1/1000 = 141/1000 = 0.141

The factors of 141 are: 1 3 47 141

141-135 = 6

141 = 10001101