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When a 4-digit number is multiplied by 1.5, the order is reversed. What is the number?

I did several things. Here is my work.

$1.5(1000a+100b+10c+d)=1000d+100c+10b+a$

$1500a+150b+15c+1.5d=1000d+100c+10b+a$

$1499a+140b=998.5d+85c$

I know $d$ must be even, or we get a decimal. Also, $d\neq0$

Now up to this point I cannot deduce the relationship of $a$ and $d$. I wrote a program to find the number but I want an intuitive approach to teach this to students.

Matthew Tang
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    Well, one thing you know is that the number is at most $6666$, for otherwise after multiplication it would have more than $4$ digits. This tells you $a<7$. – Fimpellizzeri Mar 09 '17 at 03:40

1 Answers1

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You already have $$1499a+140b=998.5d+85c$$ Multiplying the both sides by $2$ gives $$2\times 1499a+2^3\times 5\times 7b=1997d+2\times 5\times 17c$$

Setting $d=2d'$ where $d'=1,2,3,4$ and dividing the both sides by $2$ give

$$1499a+2^2\times 5\times 7b=1997\times d'+5\times 17c\tag1$$

Note here that we have to have $a\le 6$ since $7000\times 1.5=10500\gt 9999$. (Fimpellizieri has already pointed that out in a comment.)

So, in mod $5$, $$a\equiv 3d'\implies (d',a)=(1,3),(2,1),(2,6),(3,4),(4,2)$$

  • For $(d',a)=(1,3)$, $(1)\iff 28b=17c-500\lt 0$

  • For $(d',a)=(2,1)$, $(1)\iff 17c=28b-499\lt 0$

  • For $(d',a)=(2,6)$, $(1)\iff 28b=17c-1000\lt 0$

  • For $(d',a)=(3,4)$, $(1)\iff 28b=17c-1$. Since $c\equiv 1\pmod 4$, we have $c=1,5,9$ giving $b=\frac{4}{7},3,\frac{38}{7}$ respectively.

  • For $(d',a)=(4,2)$. $(1)\iff 17c=28b-998\lt 0$

Hence, $\color{red}{(a,b,c,d)=(4,3,5,6)}$ is the only solution.

mathlove
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  • Thanks, very explanatory. However, could you please elaborate on how you deduced $\mod 5$ to conclude $a\equiv3d'$ and how did you conclude $c\equiv1 (\mod4)$ for $(d',a)=(3,4)$? – Matthew Tang Mar 09 '17 at 06:15
  • @MatthewTang: In mod $5$, we have $1499\equiv -1,2^2\times 5\times 7\equiv 0,1997\equiv 2$ and $5\times 17\equiv 0$, so we get, from $(1)$, $-a\equiv 2d'$, i.e. $a\equiv -2d'\equiv 3d'$. Also, for $(d',a)=(3,4)$, in mod $4$, we have $28\equiv 0$ and $17\equiv 1$, so we get $0\equiv c-1$, i.e. $c\equiv 1$. I hope this helps. – mathlove Mar 09 '17 at 06:21
  • @MatthewTang: Let me add one thing : $a\equiv b\pmod c$ means that $a-b$ is divisible by $c$. You might want to see here. – mathlove Mar 09 '17 at 09:41