Digit sum of $x$ consisting of only 3,4,5,6 = digit sum of $2x$

Question

$x$ is a positive integer such that its digits can only be $3$,$4$,$5$, $6$. $x$ contains at least one copy of each of these four digits. The sum of the digits of $x$ is $900$ and the sum of the digits of $2x$ is also $900$. How many digits are there in the maximum value of $x$?

$3\times2=6\\4\times2=8\\5\times2=10\\6\times2=12$

Tens digit can be maximum $1$.

If in $x$ there was ___ $\rightarrow$ upon multiplication in $2x$, ___

$3\rightarrow6, \space\space3\rightarrow7\\4 \rightarrow 8, \space\space4\rightarrow9 \\5 \rightarrow 0, \space\space 5 \rightarrow1\\6\rightarrow2, \space\space 6\rightarrow3$

@XcoderX please check my comment on your answer at the very end. — Soha Farhin Pine, Jan 24 '18 at 12:04

score 1 · Accepted Answer · answered Jan 24 '18 at 11:07

1

First, we have to establish the basics:

3 multiplied 2 gives 6.

6 multiplied 2 gives 12, which has a sum of 3.

Note that $6+3=3+6$ ---> sum does not change.

4 multiplied 2 gives 8.

5 multiplied 2 gives 10, which has a sum of 1.

Note that $4+5=8+1$ ---> sum does not change.

Observation: Since the sum of digits are the same for $x$ and $2x$, we can conclude that there are equal number of 3s and 6s, and equal number of 4s and 5s, such that the sum remains the same.

Assume we have $m$ number of 3, and $n$ number of 4. Then we have $3m+6m+4n+5n=900$. $m+n=100$.

Now we want to find $2m+2n$, which is the number of digits, which is $200$. Do note that the smallest $x$ has also $200$ digits.

answered Jan 24 '18 at 11:07

QuIcKmAtHs

1,451

$6666444 \times 2 = 13332888$ both with digit sums of $36$ though $6666444$ does not have equal numbers of $3$s and $6$s, or of $4$s and $5$s – Henry Jan 24 '18 at 11:12
They have equal number of digits @Henry – QuIcKmAtHs Jan 24 '18 at 11:13
You said Observation: Since the sum of digits are the same for $x$ and $2x$, we can conclude that there are equal number of 3s and 6s, and equal number of 4s and 5s, such that the sum remains the same and I suggested $x=6666444$ as a possible counterexample – Henry Jan 24 '18 at 11:16
No. $x$ and $2x$ have equal number of digits, but not in your example – QuIcKmAtHs Jan 24 '18 at 11:17
Thanks. It's super-intuitive. That's how I had planned to solve it too. – Soha Farhin Pine Jan 24 '18 at 11:18
Please accept and upvote @SohaFarhinPine – QuIcKmAtHs Jan 24 '18 at 11:18
I do not see $x$ and $2x$ having equal numbers of digits in the question, but even if they did, then $4446666 \times 2=8893332$ would be a counterexample to your statement – Henry Jan 24 '18 at 11:21
There is at least one of each digit – QuIcKmAtHs Jan 24 '18 at 11:30
@Henry your counter argument is wrong – QuIcKmAtHs Jan 24 '18 at 11:40
But my point is not wrong. Try $34444566666$ – Henry Jan 24 '18 at 11:49
That holds true, but the main thing here is the sum 900 is a multiple of 9 – QuIcKmAtHs Jan 24 '18 at 11:51
You also have, $6\rightarrow2, 3\rightarrow7\ 6+3=2+7$ and $4\rightarrow9, 5\rightarrow0\ 4+5=9+0$ – Soha Farhin Pine Jan 24 '18 at 12:01

Digit sum of $x$ consisting of only 3,4,5,6 = digit sum of $2x$

1 Answers1