Some Special Three-Digit Numbers

First of all, note that $N$ can only start with 2,4,6 or 8. However, $N$ cannot start with 4 as $3N$ would be in the region of 1200 to 1499 which has '1' in it. For $N$ starting with 2, we have

1) $600 \leq 3N <700$ which yields $N = 200,202,208,220,222,228$ or

2) $800\leq 3N < 900$ which yields $N= 268,280,282,288$ . There are 10 cases so far.

For $N$ starting with 6, we are looking at $2000 \leq 3N<2100$ where $N=668,688,680,682$ (4 cases).

For $N$ starting with 8, we have

1) $2400\leq 3N<2500$ which yields $N=800,802,808,820,822,828$ or

2) $2600\leq 3N<2700$ , we have $N= 868,880,882,888$ (10 cases).

So there are a total of 24 numbers.

total three digit numbers= 900, whereas 500 numbers are odd numbered HUNDREDs. all 100s,300s,500s,700s &900s are left aside.

that leaves us with 400 numbers. half of these are the numbers which contains odd number in TENs place leaving us with 200 numbers. another 100 numbers are those in which UNIT digit is an odd number. finally there are a total of 100 numbers with all three even digits.

N=100

even numbers= 0,2,4,6,8

3N:

thrice the even digit itself is an even number(0,6,12,18,24)

3 times of 4 will always result in an odd carry i.e 1, therefore all those numbers that contain 4 don't qualify.

3*6=18 also results in odd carry. therefore, all those numbers with 6 in UNIT space don't meet the criterion.

numbers with 6 in TEN and/or HUNDRED place will result an even product only if 6 receives an even carry from UNIT or TEN respectively. that's possible only if there's 8 in UNIT's and/or TEN's place

in a three digit number ,0 can never appear in HUNDRED's place (obviously). it can take UNIT and/or TEN's place

2*3=6 , 2 can take any of three places

3*8=24 , 8 can appear in any of three places

> 0,2,8 qualify for UNIT place

    >for U=0 , only 0,2,8 can appear in TEN's pace
    >for U=2, only  0,2,8 can appear in TEN's pace
    >for U=8, only  0,2,6,8 can appear in TEN's pace

        >  for T=0, only 2,8 can appear in HUNDRED's space 
        >  for T=2, only 2,8 can appear in HUNDRED's space
        >  for T=6 (only if U=8), only 2,6,8 can appear in HUNDRED's space
        >  for T=8, only 2,6,8 can appear in HUNDRED's space

                   U=0, T=0,H=2/8   {two numbers}
                   U=0, T=2,H=2/8   {two numbers}
                   U=0, T=8,H=2/6/8    {three numbers}  

                     U=2, T=0,H=2/8   {two numbers}
                     U=2, T=2,H=2/8   {two numbers}
                     U=2, T=8,H=2/6/8    {three numbers} 

                        U=8, T=0,H=2/8   {two numbers}
                        U=8, T=2,H=2/8   {two numbers}
                        U=8, T=6,H=2/6/8    {three numbers}

U=8, T=8,H=2/6/8 {three numbers}

thers is a total of 24 three digit numbers with all three even digits for which thrice the number (3N) also results in a number that contains all even digits

Assume $n=\overline{abc}, 3n=\overline{defg}$ with $a,b,c,d,e,f,g$ even digits, $a$ non-zero. If $c=4,6$ then $r=1$ and $3b+1\equiv1\pmod2$ hence, not valid. If $c<4$ then $g=3c$ since $3c<10$ and no carry. $300a+30b+c\leq 2(1200+120+2)$ so $d=0,2$ with $2$ iff $a\ge6$ . Also, $a \neq4$ for $4\times3=12$ giving an odd digit.

Case 1: $d=0,a=2$ . If $c<4$ then we are reduced to $60+3b=10e+f$ which gives $6$ solutions. If $c=8$ then, $g=4$ and we have $60+3b+2=10e+f$ having $4$ solutions.

Case 2: $d=2,a=6$ If $c<4$ we are reduced to $180+3b=100d+10e+f$ which yields $2$ solutions. If $c=8$ then, $180+3b+2=100d+10e+f$ giving $2$ more solutions.

Case 3: $d=2,a=8$ If $c<4$ we have $240+3b=100d+10e+f$ which has $6$ solutions. Else, we get $240+3b+2=100d+10e+f$ which has $4$ solutions.

In total, we have $24$ solutions.

Only 0, 8 and 2 when multiplied by 3 gives an even carry over. So 2x3x3 such numbers. But is 6 is succeeded by 8, there are 3 + 3=6 such numbers that gives even digit 3N. So 24

Consider the 1-digit version of this problem. $3 \times 0 = 0, 3 \times 2 = 6, 3 \times 4 = 12, 3 \times 6 = 18, 3 \times 8 = 24$ . Hence there are 2 solutions.

Consider the 2-digit version of this problem. $N=\overline{ab}$ . The last digit has to be $0, 2, 8$ .

Case 2A: $b=0, 2$ . Then, we are back to the 1-digit version, which has 2 solutions. Thus, there are $2\times 2 = 4$ solutions here.

Case 2B: $b=8$ . Then, $3a+2$ is an even-digit number, which has $3$ solutions.

Thus, there are 7 solutions.

Consider the 3-digit version of this problem. $N= \overline{abc}$ . The last digit has to be $0, 2, 8$ .

Case 3A: $c=0, 2$ . Then we are back to the 2-digit version, which has 7 solutions. Thus, there are $2 \times 7 = 14$ solutions here.

Case 3B: $c=8$ . Then $30a+3b+2$ is an even digit number. Case 3B1: $b=0, 2$ , then $3a$ is an even digit number, which has 2 solutions. Thus there are $2 \times 2 \times 1 = 4$ solutions here. Case 3B2: $b=6,8$ , then $3a+2$ is an even digit number, which has $3$ solutions. Thus there are $3 \times 2 \times 1 = 6$ solutions here.

Hence, in total, there are $14+4+6 = 24$ solutions.