There are two number a and b. both are four digit numbers . a is reverse of b and a is four time of b.

a = reverse(b)

AND

a = 4*b

how to find a and b? with explaination

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- Jun 8th 2013, 09:54 AMashoksinghSimple Math problem
There are two number a and b. both are four digit numbers . a is reverse of b and a is four time of b.

a = reverse(b)

AND

a = 4*b

how to find a and b? with explaination - Jun 8th 2013, 01:52 PMHallsofIvyRe: Simple Math problem
Write the digits of a as "pqrs". The a= 1000p+ 100q+ 10r+ s and b= 1000s+ 100r+ 10q+ p. a= 4b becomes 1000p+ 100q+ 10r+ s= 4(1000s+ 100r+ 10q+ s).

If p, q, r, and s could be any real numbers there would be an infinite number of solutions. But here we know that p, q, r, and s must be integers from 0 to 9 (and, presumably, neither p nor s are 0). - Jun 8th 2013, 05:22 PMSorobanRe: Simple Math problem
Hello, ashoksingh!

Quote:

$\displaystyle \text{There are two four-digit numbers, }a\text{ and }b.$

. . $\displaystyle a\text{ is the reverse of }b.$

. . $\displaystyle a\text{ is four times }b.$

$\displaystyle \text{Find }a\text{ and }b.$

$\displaystyle \text{Let }\,\begin{Bmatrix}a &=& DCBA \\ b &=& ABCD \end{Bmatrix}$

$\displaystyle \text{We have an "alphametic":}$

. . $\displaystyle \begin{array}{cccc}_1&_2&_3&_4 \\ A&B&C&D \\ \times &&& 4 \\ \hline D&C&B&A \end{array}$

$\displaystyle \text{In column-1 }(c_1)\text{, we see that }4A \,=\,D$.

$\displaystyle \text{Since }D\text{ is a digit, }A\text{ must be }1\text{ or }2$.

$\displaystyle \text{In }c_4\text{, we see that }D\times 4\text{ ends in }A.$

. . $\displaystyle \text{Hence, }A\text{ must be even: }\:A = 2.$

$\displaystyle \text{We have:}$

. . $\displaystyle \begin{array}{cccc}_1&_2&_3&_4 \\ 2&B&C&D \\ \times &&& 4 \\ \hline D&C&B&2 \end{array}$

$\displaystyle \text{In }c_1\text{, we see that }D = 8\text{ or }9.$

$\displaystyle \text{In }c_4\text{, we see that }D\times 4\text{ ends in }2.$

. . $\displaystyle \text{Hence: }\,D \,=\,8$

$\displaystyle \text{We have:}$

. . $\displaystyle \begin{array}{cccc}_1&_2&_3&_4 \\ 2&B&C&8 \\ \times &&& 4 \\ \hline 8&C&B&2 \end{array}$

$\displaystyle \text{In }c_2,\,4B+ \text{(carry)}\,=\,C.$

. . $\displaystyle \text{Hence, }B\text{ must be }0\text{ or }1.$

$\displaystyle \text{If }B = 0\text{, then }c_3\text{ says: }\,4C + 3\text{ ends in }0.$

. . $\displaystyle \text{This is impossible, so }\,B = 1.$

Finally, we have:

. . $\displaystyle \begin{array}{cccc} 2&1&7&8 \\ \times &&& 4 \\ \hline 8&7&1&2 \end{array}$

- Jun 9th 2013, 07:44 AMsakonpure6Re: Simple Math problem
I can do a C++ program for you to solve the problem if you want?!

- Jun 9th 2013, 04:43 PMtopsquarkRe: Simple Math problem