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# what is 1190 divisible by

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what is 1190 divisible by

Guest Sep 23, 2014

#2
+19506
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what is 1190 divisible by

$$\small{\text{ prime factorization:  1190 = 2^{{1}}*5^{{1}}*7^{{1}}*17^{{1}}  }}$$

$$\small{\text{ The number of the divisors  = ({1} + 1) *({1} + 1) *({1} + 1) *({1} + 1) = 2*2*2*2 = 16 }}$$

$$\begin{array}{|r|c|c|c|c|c|l|l|} \hline n & 1 & 2 & 5 & 7 & 17 & & devisor\\ \hline 1 & x & & & & & &= 1 \\ 2 & & x & & & & &= 2 \\ 3 & & & x & & & &= 5 \\ 4 & & & & x & & &= 7 \\ 5 & & & & & x & &=17 \\ 6 & & x & x & & & 2*5 &= 10 \\ 7 & & x & & x & & 2*7 &= 14 \\ 8 & & x & & & x & 2*17 &= 34 \\ 9 & & & x & x & & 5*7 &= 35 \\ 10 & & & x & & x & 5*17 &= 85 \\ 11 & & & & x & x & 7*17 &= 119 \\ 12 & & x & x & x & & 2*5*7 &= 70 \\ 13 & & x & x & & x & 2*5*17 &= 170 \\ 14 & & x & & x & x & 2*7*17 &= 238 \\ 15 & & & x & x & x & 5*7*17 &= 70 \\ 16 & & x & x & x & x & 2*5*7*17 &= 1190 \\ \hline \end{array}$$

heureka  Sep 24, 2014
#1
+3450
+8

Well, 1190 is divisible by a couple things.

1190 is divisible by 119 and 10.

1190 is also divisible by 17 and 70.

#2
+19506
+8

what is 1190 divisible by

$$\small{\text{ prime factorization:  1190 = 2^{{1}}*5^{{1}}*7^{{1}}*17^{{1}}  }}$$

$$\small{\text{ The number of the divisors  = ({1} + 1) *({1} + 1) *({1} + 1) *({1} + 1) = 2*2*2*2 = 16 }}$$

$$\begin{array}{|r|c|c|c|c|c|l|l|} \hline n & 1 & 2 & 5 & 7 & 17 & & devisor\\ \hline 1 & x & & & & & &= 1 \\ 2 & & x & & & & &= 2 \\ 3 & & & x & & & &= 5 \\ 4 & & & & x & & &= 7 \\ 5 & & & & & x & &=17 \\ 6 & & x & x & & & 2*5 &= 10 \\ 7 & & x & & x & & 2*7 &= 14 \\ 8 & & x & & & x & 2*17 &= 34 \\ 9 & & & x & x & & 5*7 &= 35 \\ 10 & & & x & & x & 5*17 &= 85 \\ 11 & & & & x & x & 7*17 &= 119 \\ 12 & & x & x & x & & 2*5*7 &= 70 \\ 13 & & x & x & & x & 2*5*17 &= 170 \\ 14 & & x & & x & x & 2*7*17 &= 238 \\ 15 & & & x & x & x & 5*7*17 &= 70 \\ 16 & & x & x & x & x & 2*5*7*17 &= 1190 \\ \hline \end{array}$$

heureka  Sep 24, 2014