Divisible number.

[bigbill]

Apologies for wording the problem last week incorrectly. Here is the correct form. It was a long night !
A no started with a four. If you shifted that 4 to the other end, you got a 1/4 of the original number. It was the smallest number starting with a four that worked that way. What is it ? The soln , not given with the problem is 410256. The original problem didn't state how many digits were in the soln. Makes it trickier, from the wording I assumed the last digit in the soln would be four. Maybe it should be 4102564, both are divisible by 4. Thank you.

 

[JeffM]

Apologies for wording the problem last week incorrectly. Here is the correct form. It was a long night !
A no started with a four. If you shifted that 4 to the other end, you got a 1/4 of the original number. It was the smallest number starting with a four that worked that way. What is it ? The soln , not given with the problem is 410256. The original problem didn't state how many digits were in the soln. Makes it trickier, from the wording I assumed the last digit in the soln would be four. Maybe it should be 4102564, both are divisible by 4. Thank you.

\(\displaystyle Given:\ \dfrac{(4 * 10^n) + z }{4} = 10z + 4.\ Find\ integers\ n\ and\ z\ such\ that\ n > 0\ and\ 0 \le z < 10^n.\)

\(\displaystyle \dfrac{(4 * 10^n) + z }{4} = 10z + 4 \implies 4 * 10^n + z = 40z + 16 \implies 39z = 4 * 10^n - 16 \implies z = \dfrac{4 * 10^n - 16}{39}.\)

\(\displaystyle n = 1 \implies z = \dfrac{40 - 16}{39} = \dfrac{24}{39} = 0 + \dfrac{24}{39},\ not\ an\ integer.\)

\(\displaystyle n = 2 \implies z = \dfrac{400 - 16}{39} = \dfrac{384}{39} = 9 + \dfrac{33}{39},\ not\ an\ integer.\)

\(\displaystyle n = 3 \implies z = \dfrac{4,000 - 16}{39} = \dfrac{3,984}{39} = 102 + \dfrac{6}{39},\ not\ an\ integer.\)

\(\displaystyle n = 4 \implies z = \dfrac{40,000 - 16}{39} = \dfrac{39,984}{39} = 1,025 + \dfrac{9}{39},\ not\ an\ integer.\)

\(\displaystyle n = 5 \implies z = \dfrac{400,000 - 16}{39} = \dfrac{399,984}{39} = 10,256 + \dfrac{0}{39},\ an\ integer.\)

\(\displaystyle 4 * 10^n + z = 4 * 100,000 + 10,256 = 410,256.\)

\(\displaystyle \dfrac{410,256}{4} = 102,564 = 102,560 + 4 = 10,256 * 10 + 4.\)

 

[soroban]

Hello, bigbill!

A number starts with a 4.
If you shift that 4 to the other end, you got a 1/4 of the original number.
It is the smallest number starting with a four that works that way.
What is the number?

The solution, not given with the problem is, 410256.

From the wording I assumed the last digit in the sol'n would be 4. .No

We have a number \(\displaystyle \text{4ABCDE.}\)

Move the 4 to the end: .\(\displaystyle \text{ABCDE4.}\)
This number is \(\displaystyle \tfrac{1}{4}\) of \(\displaystyle \text{4ABCDE.}\)

Hence: .\(\displaystyle \text{ABCDE4} \:=\:\frac{1}{4}\times\text{4ABCDE}\)

That is: .\(\displaystyle \text{ABCDE4} \times 4 \:=\:\text{4ABCDE}\)


We have this alphametic:

. . \(\displaystyle \begin{array}{cccccc} _1 & _2 & _3 & _4 & _5 & _6 \\ A&B&C&D&E&4 \\ \times &&&&&4 \\ \hline 4&A&B&C&D&E \end{array}\)


In column-6, we have: \(\displaystyle E = 6\)

. . \(\displaystyle \begin{array}{cccccc} _1 & _2 & _3 & _4 & _5 & _6 \\ A&B&C&D&6&4 \\ \times &&&&&4 \\ \hline 4&A&B&C&D&6 \end{array}\)


In column-5, we have: \(\displaystyle D = 5\)

. . \(\displaystyle \begin{array}{cccccc} _1 & _2 & _3 & _4 & _5 & _6 \\ A&B&C&5&6&4 \\ \times &&&&&4 \\ \hline 4&A&B&C&5&6 \end{array}\)


In column-4, we have: \(\displaystyle C = 2\)

. . \(\displaystyle \begin{array}{cccccc} _1 & _2 & _3 & _4 & _5 & _6 \\ A&B&2&5&6&4 \\ \times &&&&&4 \\ \hline 4&A&B&2&5&6 \end{array}\)


In column-3, we have: \(\displaystyle B = 0.\)

. . \(\displaystyle \begin{array}{cccccc} _1 & _2 & _3 & _4 & _5 & _6 \\ A&0&2&5&6&4 \\ \times &&&&&4 \\ \hline 4&A&0&2&5&6 \end{array}\)


In column-2: \(\displaystyle A = 1.\)

. . \(\displaystyle \begin{array}{cccccc} _1 & _2 & _3 & _4 & _5 & _6 \\ 1&0&2&5&6&4 \\ \times &&&&&4 \\ \hline 4&1&0&2&5&6 \end{array}\)


Therefore: .\(\displaystyle \text{4ABCDE} \:=\:410256\)