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22 votes
22 votes
Check whether 934700 is divisible by 2, 3, 4, 5, 6, and 11

User Yvanscher
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1 Answer

20 votes
20 votes

Answer:

934700 is divisible be 2 because the rule for divisibility by 2 is that the last digit of the number should end in 0,2,4,6,8 . so this number will be divisible by 2

934700 isn't divisible by 3 because the rule for 3 divisibility is that if you add the digits like here 9 + 3 + 4 + 7 + 0 + 0 = 23 the sum of their digits answer should be in 3's table that is . 3,9,12,15,18... Ans soo on

934700 is divisible by 4 as the last two digits are divisible by 4

934700 is divisible by 5 because the divisibility rule of 5 is the number should end with 5 or 0

934700 isn't divisible by 6 because the rule is that the number should be both be divisible by 2 and 3 and above we saw that 3 isn't divisible so it is not divisible by 6

934700 isn't divisible by 11 the divisibility of 11 is that you have to add the odd places and even places separately like

9 + 4 +0 =13

3+ 7 + 0 = 10

and now the rule is to find the difference between them and that difference should be divisible by 11

13-10

= 3 so 3 isn't coming in the table of 11 . Hence it is not divisible

User TPPZ
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