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505-555 Level|   Number Properties|                     
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niheil
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General rule :

To find the unit's digit- find the remainder when divided by 10.

to find the last two digits - find the remainder when divided by 100.

....and so on.

For advance theory of remainders check the Math book of gmat Club.
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niheil
What is the units digit of 13^4*17^2*29^3?

(A) 9
(B) 7
(C) 5
(D) 3
(E) 1

Units digit of \(13^4\) will be 1
Units digit of \(17^2\) will be 9
Units digit of \(29^3\)will be 9

So, Units digit of \(13^4*17^2*29^3\) will be 1 * 9 * 9 => 1

Answer will be option (E) 1
:lol:
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Here's how you do this in under 1:30:

Step 1: Rephrase the question to --- (3^4) * (7^2) * (9^3)
Step 2: Apply unit digit power rules (ANEI = A number ending in) = (ANEI 1) * (ANEI 9) * (ANEI 9)
Step 3: Multiply the ends of numbers (i.e. units digits) = 1*9*9 = 81 = units digit of 1.
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Solution



Approach and Working:

The units digit of \(13^4∗17^2∗29^3\) will be same as the units digit of \(3^4 * 7^2* 9^3\).
• Units digit of \(3^4 * 7^2* 9^3\)= Units digit of (\(3^4\))* Units digit of (\(7^2\)) * Units digit of (\(9^3\))
• = 1* 9*9= 1
Hence, the correct answer is option E.

Answer: E
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To find the units digit of \(13^4 * 17^2 * 29^3\), we only need to look at the units digits of the base numbers, which are 3, 7, and 9. So this becomes a units digit problem for \(3^4 * 7^2 * 9^3\). The units digit of powers of 3 cycles every 4: [3, 9, 7, 1], so \(3^4\) ends in 1. For 7, the units digits cycle every 4: [7, 9, 3, 1], so \(7^2\) ends in 9. Powers of 9 alternate between 9 and 1: [9, 1], so \(9^3\) ends in 9. Now multiply the units digits: \(1 * 9 = 9\), and \(9 * 9 = 81\), so the final units digit is 1. Answer: (E).
niheil
What is the units digit of \(13^4*17^2*29^3\)?

(A) 9
(B) 7
(C) 5
(D) 3
(E) 1
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