  # Time to rock and roll: The Wknd puzzle with Dilip D’Souza

How good is your multiplication? With three-digit numbers? And three times three? If you’re feeling brainy, roll the dice. How many combinations can you predict?
Updated on Nov 20, 2021 01:04 PM IST
ByBy Dilip D’Souza

You have three dice. These are the kind that have numbers, not dots, on each of the faces. Thus 1, 2, 3, 4, 5, 6.

You lay the dice down in a row. The numbers facing up form a 3-digit number. For example, 215, or 461, or 332. Let’s allow that you can turn the “6” upside-down to make a “9”. Thus you can also have 491, or 999.

Question: If you add up all the possible 3-digit numbers you can form with these three dice, what would the total be?

Clearly it’s too much work to list all possible 3-digit numbers and add them up, so can you find another way to approach this puzzle?

Hint: Each dice can show one of 7 different digits. Thus if you had just one dice, you could form these 7 numbers: 1, 2, 3, 4, 5, 6 and 9. Their sum is 30.

If you had two dice, for each digit on the first, there are the 7 choices above for the second die. For example, if you chose 1 for the first die, you’d have these numbers 11, 12, 13, 14, 15, 16, 19. So you can form 49 different numbers.

Suppose you did list all 49 possibilities. Each digit appears 7 times in the units place, which means that sum of 30 (1+2+3+4+5+6+9) appears 7 times, for a total of 210 (30x7). Similarly, each digit appears 7 times in the tens place, meaning that sum of 30 appears 7 times there too. But when a digit appears in the tens place, it’s worth ten times its face value (eg, 25 is actually 20+5). So the sum in the tens place is actually 2100 (30x10x70).

And thus the total of all possible 2-digit numbers is 30x10x70 + 30x7, or 2100 + 210, or 2310.

Can you extend this logic to 3 dice?

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For each digit on the first die (the hundreds place), there are the 49 choices above for the second and third dice (tens and units places). Which means that each digit appears 49 times in each place; or, that the sum of 30 appears 49 times in each place.

So the total of all possible 3-digit numbers is 30x100x49 + 30x10x49 + 30x49, or 147000 + 14700 + 1470, or 163,170.

Left for you: What if you had 4 dice? 5?

Harder: What if you were not allowed to repeat digits? (Thus no 332, or 999).

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