What are the last two digits (at tens place and units place) in the

Question

What are the last two digits (at tens place and units place) in the numerical value of 49^852 + 11^77 ?

A. 52
B. 60
C. 72
D. 82
E. 92

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gchandana · Accepted Answer

Let's take 49^852

49^852 = 7^1704 = (7^4)^426, now 7^4 ends with 01, so 01^426 also gives 01 in the tens and units place.

11^77, let's check for a pattern
11^2 = ends in 21
11^3 = ends in 31
11^4 = ends in 41, so 11^77 should end in 771 or 71 (cycle length is 10, as in after every 11^10, the cycle repeats, from first, 11, 21, 31, 41, and so on. 77th is nothing but, after 70 cycles, 7th term will end in 71)

So, 01 + 71 = 72 is what will be in the tens and units place.
Option C

Hope it helps.

Dereno · Answer

Let’s understand the powers of number and last two digits.

Try to bring the number^exponent , such that the UNIT DIGIT is 1.

3^4 = 81

9^2 = 81

7^4 = 2401.

With the numbers ending in 1. Let’s assume the number to be HT1^n , where n is an exponent.

1^n , the unit digit must be 1.

The Tens digit will be (T*n).

49^ 852

= (7^2)^852

= 7^ 1704

=(7^4) ^426

= (2401)^426

Unit digit = 1^426 = 1.

Tens digit = (0*426) = 0

Therefore, the last two digits = 01.

11^77

Unit digit = 1

Tens digit = (77*1) = 7

Therefore, the last two digits = 71.

Hence, the sum is 01+71 =  72

Option C

Jalmagro · Answer

Find the pattern 11^2 = 121, 11^3= 1331... etc 11's tens digit goes 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. Unit digit stays at 1 49 = 7^2, 7^2(852) = 7^1704. 7's unit digit goes 7, 9, 3, 1. Tens digit is a bit trickier, but still follows a pattern: 7^2 = 49, 7^3 = 2301... tens digit will always be 0 11 = 71, 7 = 01 Add together to get 72 Answer C

ExpertsGlobal5 · Answer

Explanation:

https://www.youtube.com/watch?v=_FvKcqlzpBM

pappal · Answer

in such questions we need to check for the repetition of the patterns.
49^852+11^77
1. For 49^852=7^1704 since 1704 is divisible by 4 so last digit for certain is 1.(cyclicity 7931)
now since 7^1=07, 7^2=49,7^3=343,7^4=2401, 7^5=16807
means  the tens digit follows the pattern: 0 followed by two 4's followed by two 0's and so on. 
i.e.0 44 00 44 00.......
leaving the first zero, 44 are at the odd location and 00 are at even location when paired.
1703/2=851 (denotes for the odd location two 4's) and the remainder is 1 (denotes 0 )
hence tens place digit should be 0 or  the last two digits are 01. 
2. For 11^77 units digit is 1 for certain.
11^1=11
11^2=121
11^3=1331
11^4=14641 to conclude that  units digit of the power should be the tens digit of the value .
so last two digits are 71
hence last two digits of 49^852+11^77 are  01+71=72 C

What are the last two digits (at tens place and units place) in the

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What are the last two digits (at tens place and units place) in the

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