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Let check one by one
6 be 3 times=6,6,2,3,4,5,5,4,1,1,3,6 least repeated number=2
5 be 3 times =5,5,1,2,3,4,6,6,2,3,4,5 Least number repeated=1
Hence answer A
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Answer: A

Total numbers on Die = 6
Let X1 ,X2 ,X3 ,X4 ,X5,X6 represent each number of the dice.
At 11th throw 5 of 6 numbers would
have come twice and one number only once.
In 12th throw, one of 5 numbers comes 3rd times.

Per prompt the sum of all numbers in 12 throws:
3X6 +2X1 +2X2 +2X3 +2X4 +X5 =46

2(X1 +X2 +X3 +X4 +X5 +X6)+X6 –X5 =46
sum of all the sides of dice =21

X1 +X2 +X3 +X4 +X5 +X6=21

substituting in above.
2*21+ X6-X5= 46
X6-X5= 46-42=4

6 and 2
5 and 1 are the only possibilities.

Hence A.
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for a total of 46 which is attained
let max of 3 throws of digit be 6 ; so we get 6*3 ; 18
left with 46-18 ; 28
using given answer options ;
total of 28 can be made by 2*13 +2 ; i.e ( 13 ; 1+3+5+4)
IMO A: 2 is least value which would had turned up...

A person kept rolling a regular die until one of the numbers appeared third time on the top. This happened in 12th throw and the sum of all the numbers in 12 throws was 46. Which number could have appeared the least number of times?
A.2
B.3
C.4
D.5
E.6
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A person kept rolling a regular die until one of the numbers appeared third time on the top. This happened in 12th throw and the sum of all the numbers in 12 throws was 46. Which number could have appeared the least number of times?
A.2
B.3
C.4
D.5
E.6

The equation can be framed as :-
\(S_{9}\)+3m = 46

The other 9 rolls can not have any number rolled thrice, hence they can be rolled only in the form of 2+2+2+2+1 where we need to find that number that was rolled only once.

One of the number rolled thrice, 4 numbers rolled twice and 1 number rolled once.

3m + 2(p+q+r+s) + n = 46

3m + n = 46 - 2(p+q+r+s)

As we can see with this equation m and n both have to be either even or odd

If m and n are even then out of p, q, r, s one will be even and the other 3 will be odd => p+q+r+s = odd
If m and n are odd then out of p, q, r, s one will be odd and the other 3 will be even => p+q+r+s = odd

Rearranging equation,

\(p+q+r+s = \frac{46}{2} - \frac{(3m+n)}{2} = 23 - \frac{(3m+n)}{2}\)

Therefore \(\frac{(3m+n)}{2} = 23 - Odd Term = Even\)

For this to be even \(\frac{(3m+n)}{2}\), only m=2 and n=6 or m=6 and n=2 satisfies

Checking for m = 6, n = 2; p+q+r+s = 13
Satisfies the above equation.

Answer - A
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