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A is the set of seven consequtive two digit numbers, none of these being multiple of 10. Reversing the numbers in set A forms numbers in set B. The difference between the sum of elements in set A and those in set B is 63. The smallest number in set A can be :

I tried to write some sets and reverse them and calculate their value, but I am not able to arrive at the answer.

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Let $A = \{10a+b,10a+b+1,10a+b+2,10a+b+3,10a+b+4,10a+b+5,10a+b+6\}$, where $a \in \{1,2,\ldots,9\}$ and since none of them is divisible by $10$, we have that $b\in\{1,2,3\}$. Then $$B =\{10b+a,10b+a+10,10b+a+20,10b+a+30,10b+a+40,10b+a+50,10b+a+60\}$$ Sum of elements in $A$ is $70a+7b+21$ and sum of elements in $B$ is $70b+7a+210$. We are given $$\vert 63(a-b) - 189 \vert = 63 \implies \vert (a-b) - 3 \vert = 1$$

If $b=1$, we get that $\vert a-4\vert = 1 \implies a=3,5$.

If $b=2$, we get that $\vert a-5\vert = 1 \implies a=4,6$.

If $b=3$, we get that $\vert a-6\vert = 1 \implies a=5,7$.

The smallest number in $A$ is obtained by choosing $a=3$ and $b=1$.

Hence, the smallest number in the set $A$ is $31$.

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    $\begingroup$ Not leaving OP much to do. $\endgroup$ – Gerry Myerson Jun 28 '12 at 3:51
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Let the smallest number in $A$ be $10r+s$ where $1\le r\le9$ and $1\le s\le3$ (can you see why these restrictions apply?). Then work out the other numbers in $A$, their sum, the numbers in $B$, their sum, and the difference between the two sums. Set that to 63, and see what it tells you about $r$ and $s$.

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Write $A=\{10x+y,10x+y+1,\cdots,10x+y+6\}$. The condition of none of these being multiple of $10$ implies $y=1$, $2$ or $3$. Then $B=\{10y+x,10(y+1)+x,\cdots,10(y+6)+x\}$. The sum of the elements of $A$ is $a=70x+7y+21$ and the sum of the elements of $B$ is $b=70y+7x+210$. So, the difference is $d=63y-63x+189$. Since $d=63$, we have $y-x+3=1$, so $x=y+2$. Then, the possibilities is $\{y=1,x=3\},\{y=2,x=4\},\{y=3,x=5\}$. So the smallest number in $A$ can be $31,42$ or $53$.

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