Show that 3^n × 4^m cannot end with the digit 0 or 5 for any natural nunber n and m
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To end with 0 or 5 a number should have 5 as it prime factor
clearly 3 have only one 3 as a factor
whereas 4 = 2×2
for better understanding
(3)^n × (2×2)^m
will give you 3^n × 2^m × 2^m does not depend on what n and m is only 2 and 3 will occur as it prime factor so it will never end with 0 or 5
clearly 3 have only one 3 as a factor
whereas 4 = 2×2
for better understanding
(3)^n × (2×2)^m
will give you 3^n × 2^m × 2^m does not depend on what n and m is only 2 and 3 will occur as it prime factor so it will never end with 0 or 5
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