Will a number divisible by 3 necessarily be divisible by 6?
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Will a number divisible by 3 necessarily be divisible by 6?
To determine if a number divisible by 3 is necessarily divisible by 6, we must apply the divisibility rules for both 3 and 6:
To explore this question, let's consider a counterexample:
Take the number . The sum of its digits is , which is divisible by 3, so 9 is divisible by 3.
However, 9 is not even, so it is not divisible by 2. As a result, 9 is not divisible by 6 (because it does not satisfy the requirement to be divisible by both 2 and 3).
This counterexample demonstrates that a number divisible by 3 is not necessarily divisible by 6.
Therefore, the statement is incorrect, and the answer is No.
No
Determine if the following number is divisible by 3:
\( 352 \)
Because 6 = 2 × 3, a number must be divisible by BOTH 2 and 3 to be divisible by 6. Numbers like 9, 15, 21 are divisible by 3 but not even, so they're not divisible by 6.
Use the two-step test: First, is it even? Second, do its digits add up to a multiple of 3? Both must be true for divisibility by 6!
Sure! Try 12, 18, 24, 30, 36. Notice they're all even AND their digit sums (like 1+2=3, 1+8=9) are multiples of 3.
Odd multiples of 3 like 9, 15, 21, 27 can never be divisible by 6 because they fail the "divisible by 2" requirement. They're missing the factor of 2!
Yes! Every other multiple of 3 is divisible by 6. The pattern goes: 3 (no), 6 (yes), 9 (no), 12 (yes), 15 (no), 18 (yes)...
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