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I can't count the number of questions I've come across of this nature:
What is the units digit of (9^3)(4^5)(6^3)?
any ideas as to how I can solve questions like this?
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In these type of questions it helps to remember that all number from 1-9 when raised to a power have a cyclicity for the unit digits.
For example taking 4 for instance : 4,16,64,256.... As seen, powers of 4 repeat their unit digits at an interval of 2.
Similarly for 9 : 9,81,729,..... powers of 9 Also repeats their units digit at an interval of 2.
For 6: 6,36,216..... The units digit of powers of 6 remain the same...
Thus, for the question we get that the units digit of 9^3 = 9, 4^5 = 4, 6^3 = 6.
The multiplication of 9,4, as units digit would give 36 as the units digit and subsequent multiplication with a number with unit's digit 6 will also give 6 as the unit's digit.
TIP: Try to memorize the intervals at which the exponents of numbers from 1-9 repeat their unit's digits.
The best way to tackle such questions is following the rule of cyclicity:
consider digit 2: \(2^1= 2\) \(2^2= 4\) \(2^3= 8\) \(2^4= 16\) \(2^5= 32\) .. take a close look at \(2^5\) the units digit repeats itself. Therefore 2 has a cyclicity of 4. Similarly digits 3,7 and 8 also have a cyclicity of 4.
digits 4 and 9: cyclicity of 2, and digits 5 and 6: units digit same as number itself.
coming to the qn: (9^3)(4^5)(6^3)
Ans: 9^3: 9 has cyclicity of 2, therefore divide its index i.e 3 by 2: you get remainder as 1: Therefore units digit is 9^1 i.e 9. 4^5: 4 has cyclicity of 2, therefore divide its index i.e 5 by 2: you get remainder as 1: Therefore units digit is 4^1 i.e 4. 6^3: Units digit will be 6 itself as explained in the theory above.
Now to get the units digit of the entire expression multiply the results above : 9*4*6= 36*6= units digit of 6.
Hope my explanation is clear. Once you get hold of this method, units digit can be calculated in seconds.
Thanks Rajesh
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