There are two number a and b.

Both are 4 digit number .

a is reverse of b

a is four time of magnitude of b ( a = 4 *b)

how to find a and b ??

Explain plese

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- Jun 8th 2013, 09:58 AMashoksinghNeed Help!!! Simple math puzzle
There are two number a and b.

Both are 4 digit number .

a is reverse of b

a is four time of magnitude of b ( a = 4 *b)

how to find a and b ??

Explain plese - Jun 8th 2013, 12:13 PMemakarovRe: Need Help!!! Simple math puzzle
What have you tried?

Let a = 1000x + 100y + 10z + w where x, y, z, w ∈ {0, ..., 9}, i.e., xyyw = 4 * wyzx. (Sometimes a line is written over xyzw to show that letters stand for individual digits of a single number and to distinguish this expression from a product.)

The following will get you started. There are only two possible values for w. Why? One can easily see (without considering other digits) that the first of those values is impossible. The second one passes that test and determines x. Now, similar to the case of w, there are only two options for z. Consider both options; in each one you already know three digits.

P.S. Please don't double-post. - Jun 30th 2013, 10:44 AMSorobanRe: Need Help!!! Simple math puzzle
Hello, ashoksingh!

Quote:

$\displaystyle \text}There are two 4-digit numbers: }\,a\text{ and }b.$

$\displaystyle a\text{ is reverse of }b.\;\;a\text{ is four times the size of }b.$

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

Let $\displaystyle b \,=\,ABCD$

Let $\displaystyle a \,=\,DCBA$

We have:

. . $\displaystyle \begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\ A&B&C&D \\ \times &&&4 \\ \hline D&C&B&A \end{array}$

In column-1, we see that: $\displaystyle 4\times A \,=\,D$

. . Hence, $\displaystyle A = 1\text{ or }2.$

In column-4, we see that: $\displaystyle D\times 4$ ends in A.

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

We have:

. . $\displaystyle \begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\ 2&B&C&D \\ \times &&&4 \\ \hline D&C&B&2 \end{array}$

In column-4, $\displaystyle D \times 4$ ends in 2.

. . Hence, $\displaystyle D = 3\text{ or }8.$

In column-1, we have $\displaystyle 2 \times 4.$

. . Hence, $\displaystyle D = 8.$

We have:

. . $\displaystyle \begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\ 2&B&C&8 \\ \times &&&4 \\ \hline 8&C&B&2 \end{array}$

In column-3, we have: $\displaystyle C \times 4 + 3$ ends in $\displaystyle B.$

In column-2, we have: $\displaystyle B \times 4 + \text{(carry)} \:=\:C$

With some trial-and error, we find that: $\displaystyle C = 7,\:B = 1.$

Therefore:

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