Yes, it is divisible by 2 because it ends with 2, it is also divisible bye 3 because 1+3+2=6. 6 is divisible by 3, therefor 132 is divisible by 3.
Let three consecutive integers be n, n+1 and n+2. If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3 as these numbers will respectively leave remainders of 1 and 2. If n is not divisible by 3 then it will leave a remainder of 1 or 2. If n leaves a remainder of 1, then n+1 leaves a remainder of 2 and n+2 is therefore divisible by 3. If n leaves a remainder of 2, then n+1 is divisible by 3 and n+2 is not divisible by 3 as it leaves a remainder of 1.
Being divisible by 2 and 3: 354 (and 360 if inclusive)Trick:Numbers divisible by 2 always ends with an even number (2,4,6,8,0)So: numbers divisible by 2 strictly between 350 and 360: 352, 354, 356, 358Numbers divisible by 3 have the sum of the digits add up to 3,6,9,12 and so onSince there are only 4 numbers there, just try them one by one352: 3+5+2=10 (not divisible by 3)354: 3+5+4=12 (divisible by 3)356: 3+5+6=14 (not divisible by 3)358: 3+5+8=16 (not divisible by 3)354/2=177354/3=118
between 1 and 600 inclusive there are:300 numbers divisible by 2200 numbers divisible by 3100 numbers divisible by both 2 and 3400 numbers divisible by 2 or 3.
yes. 2 + 1 + 6 = 9 9 is divisible by 3 so 216 is divisible by 3.
Yes, it is divisible by 2 because it ends with 2, it is also divisible bye 3 because 1+3+2=6. 6 is divisible by 3, therefor 132 is divisible by 3.
Let three consecutive integers be n, n+1 and n+2. If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3 as these numbers will respectively leave remainders of 1 and 2. If n is not divisible by 3 then it will leave a remainder of 1 or 2. If n leaves a remainder of 1, then n+1 leaves a remainder of 2 and n+2 is therefore divisible by 3. If n leaves a remainder of 2, then n+1 is divisible by 3 and n+2 is not divisible by 3 as it leaves a remainder of 1.
To be divisible by 6, the number must be divisible by both 2 and 3:To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3.As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3.If the number is not divisible by 2 or 3 (or both) then the number is not divisible by 6.examples:126Last digit is even so it is divisible by 2 1 + 2 + 6 = 9 which is divisible by 3, so it is divisible by 3→ 126 is divisible by both 2 and 3, so it is divisible by 6124Last digit is even so it is divisible by 2 1 + 2 + 4 = 7 which is not divisible by 3, so it is not divisible by 3→ 126 is divisible by 2 but not divisible by 3, so it is not divisible by 6123Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 3 = 6 which is divisible by 3, so it is divisible by 3→ 123 is divisible by 3 but not divisible by 2, so it is not divisible by 6121Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 1 = 4 which is not divisible by 3, so it is not divisible by 3→ 121 is not divisible by either 2 or 3, so it is not divisible by 6
The numbers 2, 3, and 9 are all divisible by 1.
If you multiply 417 by any integer (0, 1, 2, 3, -1, -2, -3, etc.), you will get a number that is divisible by 417.If you multiply 417 by any integer (0, 1, 2, 3, -1, -2, -3, etc.), you will get a number that is divisible by 417.If you multiply 417 by any integer (0, 1, 2, 3, -1, -2, -3, etc.), you will get a number that is divisible by 417.If you multiply 417 by any integer (0, 1, 2, 3, -1, -2, -3, etc.), you will get a number that is divisible by 417.
Being divisible by 2 and 3: 354 (and 360 if inclusive)Trick:Numbers divisible by 2 always ends with an even number (2,4,6,8,0)So: numbers divisible by 2 strictly between 350 and 360: 352, 354, 356, 358Numbers divisible by 3 have the sum of the digits add up to 3,6,9,12 and so onSince there are only 4 numbers there, just try them one by one352: 3+5+2=10 (not divisible by 3)354: 3+5+4=12 (divisible by 3)356: 3+5+6=14 (not divisible by 3)358: 3+5+8=16 (not divisible by 3)354/2=177354/3=118
A number is divisible by 6 if it is divisible both by 2 and 3 1240 is even then is divisible by 2 1+2+4+0 = 7 which is not divisible by 3 then 1240 is not divisible by 3 Thus 1240 is not divisible by 6
no because: 1+4+1+2=8 and 8 is not divisible by 3.
No, it is divisible by 1, 3, 13, 39.
between 1 and 600 inclusive there are:300 numbers divisible by 2200 numbers divisible by 3100 numbers divisible by both 2 and 3400 numbers divisible by 2 or 3.
352 = (3 x 100) + (5 x 10) + (2 x 1)
1, 2, 7, 3 and 6 are not divisible by 4 and/or 9. 12736 is divisible by 4 but not by 9.