The quantity $3^34^45^56^6 – 3^64^55^46^3$ will end in how many zeros?
This is a GRE Practice question and I have to do it without using a calculator. Can anyone help me on this?
[Math] How many zeroes at the end of $3^34^45^56^6 – 3^64^55^46^3$
elementary-number-theorygre-exam
Best Answer
Using $a^m\cdot a^n=a^{m+n}; a^{mn}=(a^m)^n$ and $(a\cdot b)^m=a^m\cdot b^m$ where $a,b,m$ are real numbers
$$3^34^45^56^6 - 3^64^55^46^3$$
$$=3^3(2^2)^45^5(2\cdot3)^6 - 3^6(2^2)^55^4(2\cdot3)^3$$
$$= 2^{8+6}5^53^{6+3} - 3^{3+6}2^{10+3}5^4$$
$$=2^{13}5^43^9(2\cdot5-1)$$
As $10=2\cdot5,$ the highest power of $10$ will be min $(13,4)$