Explain why (42)n cannot end with zero. Also find the unit's place digit in the expansion of (42)n if n=17
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If you mean 42*n then it can end with zero, say for n=5
But 42^n cannot end with zero as 42=2*3*7
42^n will never have both 5 and 2 (only 2) as it’s prime factor, so 42^n will never end with zero
42^1 will end with 2
42^2 will end with 4
42^3will end with 8
42^ 4will end with 6
42^5 will end with 2
Now you will observe these no.s repeating after 4
17=4*4+1 so 14^17 will end with 2
Hope I helped you
But 42^n cannot end with zero as 42=2*3*7
42^n will never have both 5 and 2 (only 2) as it’s prime factor, so 42^n will never end with zero
42^1 will end with 2
42^2 will end with 4
42^3will end with 8
42^ 4will end with 6
42^5 will end with 2
Now you will observe these no.s repeating after 4
17=4*4+1 so 14^17 will end with 2
Hope I helped you
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if the number 42n for any n, were to end with the digits 0 it's prime factors must be 2 & 5. But it's prime factorization is 2, 3 & 7 .so, it cannot end with the digits 0.
expansion:
unit place of:
n=17
14*14+7 = 17
16+1=17
17= 17
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