How about 27
405 is divisible by: 1, 3, 5, 9, 15, 27, 45, 81, 135, 405.
27 is divisible by 1, 3, 9, 27.
No. 135 is divisible by: 1, 3, 5, 9, 15, 27, 45, 135.
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No.
How about 27
405 is divisible by: 1, 3, 5, 9, 15, 27, 45, 81, 135, 405.
27 is divisible by 1, 3, 9, 27.
No. 135 is divisible by: 1, 3, 5, 9, 15, 27, 45, 135.
27 is wholly divisible by 1, 3, 9, and 27.
The number 350 is even, so it's divisible by 2; 350/2 = 175. 175 ends with a 5, so it's divisible by 5; 175/5 = 35. The same applies to 35; 35/5 = 7. Our factors are now 2 X 5 X 5 X 7, which are all prime numbers. Therefore, that's the prime factorization of 350: 2 X 52 X 7.
Subtract 8 times the last digit from remaining truncated number. Repeat the step as necessary. If the absolute of result is divisible by 27, the original number is also divisible by 27 Check for 945: 94-(8*5)=54; 5-(8*4)=-27 Since 27 is divisible by 27, the original no. 945 is also divisible. Check for 264681: 26468-(8*1)=26460; 2646-(8*8)=2582; 264-(8*6)=216 21-(8*6)=-27 Since 27 is divisible by 27, the original no. 264681 is also divisible. Check for 81: 8-(8*1)=0; Since 0 is divisible by 27, the original no. 81 is also divisible.
1 (350/1=350) 2 (350/2=175) 5 (350/5=70) 7 (350/7=50) 10 (350/10=35) 14 (350/14=25) 25 (350/25=14) 35 (350/35=10) 50 (350/7=50) 70 (350/70=5) 175 (350/175=2) 350 (350/350=1)
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No.
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