Division with specific quotients and remainders

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Division with specific quotients and remainders

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Source: Paraguayan National Olympiad 2011, Level 3, Problem 3

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#1 h Aug 31, 2014, 12:59 AM • 2 Y

Y by Adventure10, Mango247

If number $\overline{aaaa}$ is divided by $\overline{bb}$ , the quotient is a number between $140$ and $160$ inclusively, and the remainder is equal to $\overline{(a-b)(a-b)}$ . Find all pairs of positive integers $(a,b)$ that satisfy this.

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#2 h Aug 31, 2014, 11:12 AM • 1 Y

Y by Adventure10

According to the last condition the remainder of this division $\overline{aaaa}$ by ${\overline{bb}}$ is $\overline{(a-b)(a-b)} = 11(a-b)$ . This combined with the fact that

${\textstyle \frac{\overline{aaaa}}{\overline{bb}} = \frac{1111a}{11b} = \frac{101a}{b}}$ ,

yields

$101a = qb + a-b$ ,

i.e.

${\textstyle q - 1 = \frac{100a}{b}}$ ,

implying

$(1) \;\; b|100a$ .

Also we have

${\textstyle 140 < \frac{101a}{b} < 160}$ ,

or alternatively,

$(2) \;\; {\textstyle \frac{140}{101}b < a < \frac{160}{101}b}$ .

Clearly ${\textstyle \frac{140}{101}b < a \leq 9}$ , which means ${\textstyle b < \frac{909}{140} < 7}$ . Moreover, if $b=1$ , then ${\textstyle 1 < \frac{140}{101} < a < \frac{160}{101} < 2}$ . In other words, $2 \leq b \leq 6$ . If $b \in \{2,4,5\}$ , then (1) is satisfied. If $b \in \{3,6\}$ , then (1) is satisfied iff $3|a$ . Using this fact combined with (2), the result is:

$b=2 \;\; \Rightarrow \;\; {\textstyle 2\frac{78}{101} < a < 3\frac{17}{101}} \;\; \Rightarrow \;\; a=3$ .

$b=3 \;\; \Rightarrow \;\; {\textstyle 4\frac{16}{101} < a < 4\frac{76}{101}}$ .

$b=4 \;\; \Rightarrow \;\; {\textstyle 5\frac{55}{101} < a < 6\frac{34}{101}} \;\; \Rightarrow \;\; a=6$ .

$b=5 \;\; \Rightarrow \;\; {\textstyle 6\frac{94}{101} < a < 7\frac{93}{101}} \;\; \Rightarrow \;\; a=7$ .

$b=6 \;\; \Rightarrow \;\; {\textstyle 8\frac{32}{101} < a < 9\frac{51}{101}} \;\; \Rightarrow \;\; a=9$ .

Summa summarum, there are four pair $(a,b)$ of digits satisfying the given conditions, namely

$(a,b) = (3,2), \: (6,4), \: (7,5) \: (9,6)$ .

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