That's one big number!

$N = \frac {3}{9}(10^{61} -1 )$ and $M = \frac {6}{9} (10^{62}-1)$ , so their product is

$\begin{array}{l l} N \times M & = \frac {2} {9} \times (10^{61} -1 )(10^{62} -1 ) \\ & = \frac {2}{9} (10^{61} - 1)\times 10^{62}- \frac {2}{9} (10^{61}-1) \\ & = \underbrace{22 \ldots 2}_{61 \, 2's} \underbrace{00 \ldots 0}_{62 \, 0's} - \underbrace{22 \ldots 22}_{61\, 2's}\\ & = \underbrace{22\ldots 2}_{60 \,2's} 1 9 \underbrace{77 \ldots 7}_{60 \, 7's} 8 \\ \end{array}$

Hence, the digit sum is $60\times 2 + 1 + 9 + 60\times 7 + 8 = 558$ .

Note: A student can also recognize the pattern by looking at small cases.

Note that a simple way of expressing a number with $x$ 1's is $\frac{10^x-1}{9}$ . This works because subtracting one from a power of ten gives a number with only nines, and then dividing by nine will result in only 1's. We can multiply such a number by 3 or 6 to get $N$ and $M$ . $N=\frac{1}{3}(10^{61}-1)$ and $M=\frac{2}{3}(10^{62}-1)$ . Thus, $N\times M=\frac{2}{9}(10^{123}-10^{62}-10^{61}+1)$ .

$10^{123}$ has 124 digits. Once $10^{62}$ is subtracted from it, it will have 123 digits--nines from the 123rd digit to the 62nd, zeroes from digit 61 until the end. Once we subtract $10^{61}$ , the 90 formed by the last nine and the first zero turn into an 89. Finally, we add one. Thus, our number has nines from digit 123 to digit 64, an eight at digit 63, a nine at digit 62, zeroes from digit 61 to digit 1, and a one at the units digits. Now we multiply by $\frac{2}{9}$ . The nines in the front of the number all turn into twos. We have $123-64+1=60$ twos. Now we have to multiply the remaining part, $8900...001$ . If we first try multiplying with $891$ , $8901$ , $89001$ , a pattern quickly emerges: the answer will be $1977...778$ . In our case, there will be 60 sevens. Adding up all the digits, $1+2\times60+7\times60+8+9=558$ , the answer.

[Note: The simpler way to approach the second paragraph is to take $N \times M = \frac{2}{9}(10^{61}-10^{1})\times 10^{62}-\frac {2}{9}(10^{61}-1) = \underbrace{22 \ldots 2}_{61 \, 2's} \underbrace{00 \ldots 0}_{62 \, 0's} - \underbrace{22 \ldots 22}_{61\, 2's}.$ Having made that handy dandy observation in the first paragraph, feel free to use it again to simplify your working :) - Calvin]

We let n(x) denote the number 333...333, with x 3's; we also let m(x) denote the number 666...666, with x 6's, where x is a positive integer. We let k(x) denote the following number 222...22219777...7778, with x 2's and x 7's, where similarly x is a positive integer.

Let the following statement "The product of n(x) and m(x+1) is k(x-1)" be labelled as P(x), where x is a positive integer. Clearly, P(1) is true because 3 x 66 = 198.

We wish to show that, assuming P(x) is true for a certain positive integer x, P(x+1) is also true; together with the fact that P(1) is true, the induction hypothesis will hence be complete.

Note the following: [n(x+1)][m(x+2)] = [n(x) + 3(10^x)][m(x+1) + 6(10^(x+1))] = [n(x)][m(x+1)] + [m(x+1)][3(10^x)] + [n(x)][6(10^(x+1))] + [3(10^x)][6(10^(x+1))]

By the induction hypothesis, n(x)m(x+1) = k(x-1). [3(10^x)][6(10^(x+1))] = 18(10^(2x+1)) [m(x+1)][3(10^x)] = 1999...9998000...000, with x 9's and x 0's. [n(x)][6(10^(x+1))] = 199...99980000...000, with (x-1) 9's and (x+1) 0's.

Hence, summing these four summands up we have the desired result, that n(x+1)m(x+2) = k(x); substituting x = 60 we find the digit sum of k(60) is simply (2 x 60 + 1 + 9 + 7 x 60 + 8) = 558.

I guessed there has to be a pattern if not the question would not be of such big numbers. And my hunch was right

I just tried 66 * 3 at first since as the question goes there is 1 more time the digit 6 appears in M than the number of times the digit 3 appears in N. The digit sum for 66 * 3 is 18.

When I tried 666 * 33 and 6666 * 333, the digit sums are 27 and 36 respectively.

From these 3 examples, one can see the common difference is always 9 when the number of times the digit 6 and 3 appear increases by one.

So, it has to do with the 9 times table, and one can easily see the number of times the digit 6 appears multiplied by 9, would give the digit sum.

In the case of 62 6s, the answer would be 62 * 9 which is equals to 558

Since there is 1 more digit 6 of M than N, the following multiplication will have 1 more digit 6 than 3

66x3=198 The digit sum will be 1+9+8= 18 666x33=21978 The digit sum will be 1+9+8+2+7=27 6666x333=2219778 The digit sum will be 1+9+8+2+2+7+7=36

The above pattern shows that for every increase in 1 digit of 6 and 1 digit of 3, the digit sum will increase by 9. To get the digit of sum of NxM, 18+ (61-1)x9= 18+540=558

66*3=198

666*33=21978

6666*333=2219778

66666*3333=222197778

simarly the numbers of 2's and 7's are same in every multiplication and 19 will come in between and 8 will come in last and the number of 2's and 7's are two less than the no of digits in the bigger number , so

66666..............6 * 333333333.................3 = 22222.........222219777777.......7777778 ,

then the sum of digits is ,

60 x 2 + 1 + 9 + 60 x 7 + 8 = 558

we perform a bit pattern and we can notice that: 3 66=198, 33 666=21978, 333 6666=2219778, . . . then we conclude that 333…33(61 3's) 666.....66(62 6's)=22....22(60 2's)1977....77(60 7's)8. and the generalization:333…33(x 3's) 666.....66(x+1 6's)=22....22(x-1 2's)1977....77(x-1 7's)8. hence the digit sum is 2 60+7*60+1+9+8=558

The numbers given in the following sequence follow a pattern .

$666\times 33$ = 21978 $6666\times 333= 2219778$

Similarly,

$66.......6\times 33..........3 = 22....21977....78$ Where if there are 'n' 3's and n+1 6's then the solution contains n-1 2's and n-1 7's

Here n = 61 , therefore the digital sum =

$60\times 2 + 1 + 9 + 60\times 7 + 8) = 120 + 18 +420 = 558$

I did something that worked fine. Notice that if we have $n$ amount of $6's$ then the digit sum is $9*n$ .

This worked for lower cases so I simply got $9*62 = 558$ :)

If you multiply 2 numbers both consist entirely of 3's but one has one more digit than the other, the sum of digit of the result is the sum of digit of the larger one x 3. i.e. number of digit of the large one x 9.

For example: 333 x 33 = 10989 , the sum of digit is 27 which is 3x9. Another example: 3333 x 333 = 1109889, the sum of digits is 36 which is 4x9 . This pattern remains the same when multiplied by 2. In example1: 2 x 10989 = 21978 the sum of digit is still 27. In example 2: 2 x 1109889 = 2219778 the sum of the digits is still 36.

In our question we have: 333....333 (61 times) x 666 ....666 (62 times) = 333...333 (61 times) x 2(333....333) {62 times} Hence the sum of the digits of the result is 62 x 9 = 558.

Digit sum of ( 6 * N ) = digit sum of ( 6 x 333..........3) = 8 + 9 * 60 +1 = 549

Digit sum of (N * M) = 62 * 549 /61 = 558

The best Way Is INDUCTION

We will observerve sum of digits

1)3x66=18

2)33x666=27

i.e. 3333...(n-1)x666....(n)=9xn

therefore

9x62=558