31 of 100: Some Sum Riddle

The sum is 133.

Simply factor 1000 into 5x5x5x2x2x2. Once a 5 and 2 are used to form a factor a zero appears since the number will be a multiple of 10. That leaves the single possibility of 8 and 125 that sums to 133.

The real question is: why is this puzzle rated more than one cactus?

The factors of $1000$ are

$1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000$

Notice that all of the factors containing the digit $0$ are multiples of $10$ . Now, any multiple of $10$ is of the form $2^a \cdot 5^b$ for positive integers $a$ and $b$ . Thus, since neither of the numbers contain the digit $0$ , they cannot be products of any permutation of $2$ and $5$ . This extends to the one-digit factors that are paired with factors that are multiples of $10 -$ namely, $2 \ (\times 50{\color{#3D99F6}{0}})$ , $4 \ (\times 25{\color{#3D99F6}{0}})$ , and $5 \ (\times 20{\color{#3D99F6}{0}})$ .

However, since $1000$ is a multiple of $10$ , it must necessarily be of the form $2^a \cdot 5^b$ . Indeed, its prime factorization is $2^3 \cdot 5^3$ , or $8 \times 125$ . Recall that $10n = 2^a \cdot 5^b$ , where $a$ and $b$ are positive integers. We can rewrite $1000$ 's prime factorization as $(2^3 \times 5^0) \cdot (2^0 \times 5^3)$ . Neither number is a multiple of $10$ , instead being only powers of $2$ or powers of $5$ , respectively.

Hence, our numbers are $8$ and $125$ , and the desired sum is $\boxed{133}$ .

Because every number with a 0 is divisible by both 2 and 5, we start by factoring 1000. $1000 = 500 * 2 = 250 * 4 = 125 * 8 => 125 + 8 = 133$ , which is our desired answer.

Let the two numbers be $m$ and $n$ . Thus we have $mn= 1000 = 2^3 \cdot 5^3.$ Obviously one of these must be even, say $m$ is even. If $m$ contains any power of $5$ , then $m$ will be a multiple of $10$ , which is not allowed. Thus it must be $n$ that is divisible by $5^3$ . Now if $n$ is even, then it will be a multiple of $10$ too, which is not allowed. So all of $2^3$ must be contained in $m$ . Therefore $2^3 \mid m$ $5^3 \mid n$ but $mn = 1000$ . It follows that $m = 2^3 = 8$ and $n = 5^3 = 125$ , so $m+n = \boxed{133}$ .

By listing all pairs of positive integers that multiply to $1000$ ,

$\begin{aligned} 1000 \ &= \ \ \ 1 \times 1000 \\ &= \ \ \ 2 \times 500 \\ &= \ \ \ 4 \times 250 \\ &= \ \ \ 5 \times 200 \\ &= \ \ \ 8 \times 125 \\ &= \ 10 \times 100 \\ &= \ 20 \times 50 \\ &= \ 25 \times 40 \end{aligned}$

we find that $\left\{8, 125\right\}$ is the only pair satisfying the condition.

Hence, the sum is $8 + 125 = \boxed{133}$ .

Numbers divisible by 10 are divisible by 2 and 5. $1000=2^3\times5^3$ , so to make sure the numbers are not divisible by 10, we must make sure that we don't mix 2 and 5 in the same factor. Therefore, one factor must be only 2s and the other should be only 5s. This gives the two factors 8 and 125, which add together to 133.

I just guessed

The first thing came to my mind was " $125 \times 8 = 1000$ " because we learned this in school as a concept. Both 125 & 8 don't contain the digit 0. So $\boxed{125 + 8 = 133}$

factor 1000 into 5 5 5 2 2*2

do not do 5*2 because that = 10

5 5 5 and 2 2 2 do not do that

add them up

133

Make a systematic table of factor pairs

1 x 1000

2 x 500

4 x 250

5 x 200

8 x 125 <- this is it, it is the only factor pair with no zeros.

10 x 100

20 x 50

25 x 40

I agree that this problem is ridiculously easy.

I solved this quickly by recalling the first few common fractions as decimals:

$\frac{1}{2}$ = 0.5

$\frac{1}{3}$ = 0.333...

$\frac{1}{4}$ = 0.25

$\frac{1}{5}$ = 0.2

$\frac{1}{8}$ = 0.125

Factor these up and you have 2+500, 3+333.3..., 4+250, 5+200, 8+125. The only one here that really works is $\frac{1}{8}$ and hark when you add 125 and 8 you get 133 which is one of the answers

The sum is 133. You can work out the prime factors of 1000 which are: 2, 2, 2, 5, 5, 5. These factors can also be written as 2 cubed and 5 cubed - 8 and 125. These are the two numbers the question talks about. Add these two together and you get 133 - the answer to the question.

250 x 4 =1000

So, I halved 250 to get 125 and then doubled 4 to get 8; 125 x 8 =1000

125 + 8 = 133

1000 = 10 * 10 * 10 = 5 * 2 * 5 * 2 * 5 * 2 = (5 ^ 3) * (2 ^ 3) = 125 * 8 (Answer)

Note that $1000=2^35^3$ . The only factors that don't involve 0 is $2^3=8$ and $5^3=125$ , and their sum is $8+125=\boxed{133}$ .

So each number can't end in 0

x*y=1000

x+y=

x = 125

y = 8

125*8=100

125+8=133

So 133 is the answer

There is no need to factor 1000 in order to find the correct answer. Since the product of the two numbers is 1000, one of them must be divisible by 5. Since that number does not contain a zero digit, it must not be divisible by 2. Hence the other number must be even, and the sum of the two numbers must be odd. Since 1001 is too large to be the sum of two numbers whose product is 1000, that leaves 133 as the only possible answer.

1000 factors to 5^3x2^3 (5x5x5x2x2x2). Since any number that has factors of 5 and 2 has the ones digit as a zero, the 5's and 2's must stay separate. This means that the two factors must be 125 (5^3) and 8 (2^3). 125+8=133

1000 has a prime factorization of (2•2•2) • (5•5•5). Considered separately, each prime factor family contributes four factors: (2•2•2) contributes the four factors 1, 2, 4 and 8, while (5•5•5) contributes the four factors (1, 5, 25 and 125). Combined, this leads to 1000 having 16 factors: (1, 2, 4, 8) • (1, 5, 25, 125) or 8 factor-pairs. The eight factor pairs are: 1•1000, 2•500, 4•250, 5•200, 8•125, 10•100, 20•50 and 25•40. Only 8•125 is a factor-pair where neither factor contains a 0 digit. An ending 0 digit would result when a factor of 2 combines with a factor of 5. Therefore, the sum 8+125 = 133 is the answer.

BTW, methinks writing the solution for this challenge was more challenging than the actual challenge!

5 x 2 is the basic power of ten combo. So the third power of 10. 5 to the third is 125, 2 to the third is 8.

factorize 1000 that gives you 2x2x2x5x5x5. the digit mustn't have 0, so we should not multiply 5 and 2 together but separately. That gives 2x2x2=8 and 5x5x5=125

• P = a*b = 1000
• S = a + b
• we have: t^2 - S*t + P = 0 (with P =1000)
• use your calculator and replace S with 110,133,254,1001 until you get the numbers satisfying the condition
• S=133 is the answer with a=t1=125, b=t2=8
• Of course this is not the best way tho.

1. 1000=10^3=(2x5)^3=2^3 x 5^3=8x125
   Actually, it is very easy to see that
   110 is not the right answer: 100x10=1000
   because 100 and 10 contain the digit 0;
   254 is not the right answer: 250x4=1000
   because 250 contains the digit 0;
   1001 is not the right answer: 1000x1
   because 1000 contains the digit 0.

1000 = 8 * 125 125 + 8 = 133

Factoring 1000 we have $1000=2^3 \cdot 5^3$ so now we have to find a divider of 1000 that ends with no zeros.

A number ends with zero when in his factorization a $2$ and a $5$ appear, so our divider's factorization won't have a $2$ and a $5$ together.

So one number will be $2 \cdot 2 \cdot 2=2^3=8$ and the other $5 \cdot 5 \cdot 5 =5^3=125$ .

$8+125=133$

$1000 = 10 \times 10 \times 10\ = 10^3$

Factors of 10 are (10, 1) or (2, 5). We know that any factor of 1000 that is divisible by 10 must contain a zero digit, which is not allowed.

$\Rightarrow 1000 = (2 \times 5)^3 = (2^3 + 5^3)$

$\Sigma (2^3 + 5^3) = (8 + 125) = \boxed{133}$