For proof of 3 divisibility, abc is a divisible by 3 if the sum of abc (a + b + c) is a multiple of 3.
<h3>
Integers divisible by 3</h3>
The proof for divisibility of 3 implies that an integer is divisible by 3 if the sum of the digits is a multiple of 3.
<h3>Proof for the divisibility</h3>
111 = 1 + 1 + 1 = 3 (the sum is multiple of 3 = 3 x 1) (111/3 = 37)
222 = 2 + 2 + 2 = 6 (the sum is multiple of 3 = 3 x 2) (222/3 = 74)
213 = 2 + 1 + 3 = 6 ( (the sum is multiple of 3 = 3 x 2) (213/3 = 71)
27 = 2 + 7 = 9 (the sum is multiple of 3 = 3 x 3) (27/3 = 9)
Thus, abc is a divisible by 3 if the sum of abc (a + b + c) is a multiple of 3.
Learn more about divisibility here: brainly.com/question/9462805
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Answer:
Step-by-step explanation:
(10 pushups)/(25 sec) = (⅖ pushup)/sec
Or,
(25 sec)/(10 pushups) = (2.5 sec)/pushup
Answer:
-a^2 + ab
Step-by-step explanation:
(a^2 - 2ab +b^2) + (2a^2 + 2ab + b2) = 3a^2 + 2b^2
(a^2 - b^2) + (a^2 + ab + 3b^2) = 2a^2 + 2b^2 + ab
subtract the first part from the second part:
(2a^2 + 2b^2 + ab) - (3a^2 + 2b^2) = -a^2 + ab
