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ZanzabumX [31]
2 years ago
11

If the last digit of a number is 0 or 5, then the number is divisible by 10. True False

Mathematics
2 answers:
Whitepunk [10]2 years ago
3 0

Answer:

False

Step-by-step explanation:

15 isn't divisible by 10

dangina [55]2 years ago
3 0

Answer:

That's false.

Step-by-step explanation:

10 can't go in to for an example 45 or 55. But, it can go into 40 or 50

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Answer:

Step-by-step explanation:

To prove divisibility, we need to factor the divident such that one of its factors matches the divisor.

(I use the notation x|y to denote that x divides y)

(A)

75^{30}|45^{45}\cdot15^{15}\\45^{45}\cdot15^{15}=3^{45}\cdot 15^{45}\cdot 15^{15}=\\=3^{45}\cdot 15^{60}=3^{45}\cdot 15^{30}\cdot 15^{30}=3^{45}\cdot (3\cdot5)^{30}\cdot 15^{30}=\\=3^{45}\cdot 3^{30}\cdot(5\cdot 15)^{30}=3^{45}\cdot 3^{30}\cdot(75)^{30}\\\implies\\75^{30}|3^{45}\cdot 3^{30}\cdot75^{30}

(B)

72^{63}|24^{54}\cdot 54^{24}\cdot2^{10}\\24^{54}\cdot 54^{24}\cdot2^{10}=(2^{162}\cdot 3^{54})\cdot(2^{24}\cdot 3^{72}) \cdot 2^{10}\\=2^{196}\cdot 3^{126}

In this case, it is easier to also factor the divisor to primes:

72^{63}=2^{189}\cdot 3^{126}

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2^{189}\cdot 3^{126}|2^{196}\cdot 3^{126}\implies\\2^{189}|2^{196}\,\,\mbox{and}\,\,3^{126}|3^{126}

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