3
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$Given$:

A, B, C, D, P, Q, R, S, T are distinct digits varying from 1 to 9.

AB, CD, PQ, RST are all concatenated Numbers.

From the given expression , figure out all the digits to make lucky seven.

$$\large 7 = \frac{AB}{CD} + \frac{PQ}{RST} $$

No computers or calculators. Let not the simple expression fool you...good deal of right deductive reasoning is involved.

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  • $\begingroup$ Is that a digit 7 on the left? $\endgroup$ – rhsquared Jun 11 at 10:07
  • $\begingroup$ Yes...seven as in Lucky seven.. $\endgroup$ – Uvc Jun 11 at 10:14
5
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It seems there is just one solution

$7 = \frac{86}{13} + \frac{95}{247}$

Reasoning

$\frac{PQ}{RST} < 1$ which means that $\frac{AB}{CD} > 6$.
Since $AB < 100$, this means $CD \leq 16$ and so $C=1$.
Since $C =1$, this means that $R \geq 2$ and so $\frac{PQ}{RST} < \frac{1}{2}$ and so $\frac{AB}{CD} > 6 \frac{1}{2}$.
This means that $CD \leq 15$.

If $CD = 15$ then $AB > 6 \frac{1}{2} \times 15 = 97 \frac{1}{2}$ and so $AB = 98$ is the only possibility.
This means that $\frac{PQ}{RST} = 7 - \frac{98}{15} = \frac{7}{15}$.
From here notice that we need to construct a two-digit number $PQ$ from the digits $2,3,4,6,7$ which is divisible by $7$. There are just two possibilities for this, $PQ = 42$ and $PQ = 63$ which would respectively give $RST = 90$ and $RST = 135$, both invalid solutions.
This means that $12 \leq CD \leq 14$.

If $CD = 12$, then $84 = 7 \times 12 > AB > 6\frac{1}{2} \times 12 = 78$ so the only allowed possibilities for $AB$ are $79$ and $83$, as we cannot reuse $1$ or $2$.
In each case the results of taking $7 - \frac{AB}{12}$ are $\frac{5}{12}$ and $\frac{1}{12}$, respectively.
In the first, case $PQ$ must be divisible by $5$ and so $Q=5$. This means we have $PQ \leq 85$. However, this would put $RST \leq 204$ which is not permitted since, the first digit of $R$ cannot be $1$ or $2$ here.
In the second case, we must use the digits $4, 5, 6, 7$ and $9$ to construct a fraction $\frac{PQ}{RST} = \frac{1}{12}$. Given that we must then have $Q \times 12 \equiv T $ mod $10$, the only possibilties for $Q$ and $T$ are $7$ and $4$ respectively. Checking the remaining possibilities for $PQ$, we have $57 \times 12 = 684$, $67 \times 12 = 804$ and $97 \times 12 = 1164$, none of which are satisfactory solutions.
Hence $13 \leq CD \leq 14$

If $CD = 13$ then $91 = 7 \times 13 > AB > 6\frac{1}{2} \times 13 = 84 \frac{1}{2}$.
This leaves the possibilities for $AB$ as being $85, 86, 87$ and $89$.
The results of taking $7 - \frac{AB}{13}$ are $\frac{6}{13}, \frac{5}{13}, \frac{4}{13}$ and $\frac{2}{13}$, respectively, which we'll label as Cases $1.1,1.2,1.3$ and $1.4$.
In Case 1.1, we must use the digits $2, 4, 6, 7$ and $9$ to construct the fraction $\frac{6}{13}$. But since $PQ \leq 97$, this means that $RST$ would be less that $211$ and we cannot have either $R$ or $S$ being $1$ or $0$ so there is no valid solution in this case.
In Case 1.2, we must use the digits $2,4,5,7$ and $9$ to construct the fraction $\frac{5}{13}$ which means $Q=5$. In this case, if $P \leq 7$, then $R \leq 1$, which is not allowed. Hence, we must have $PQ = 95$ and, in this case we find that $RST = 95 \times \frac{13}{5} = 247$ which works with the remaining digits!
In Case 1.3, we must use the digits $2,4,5,6$ and $9$ to construct the fraction $\frac{4}{13}$. Again, since the numerator must be divisible by $4$, if $P \leq 6$, then $RST \leq 208$ which is not permitted since we cannot have $S=0$ or $R=1$. Hence, $P=9$ is the only possibility here, which means $Q=2$ or $6$ to make $PQ$ divisible by $4$.
If $PQ = 92$ then $RST = 92 \times \frac{13}{4} = 299$, which is not permitted. If $PQ=96$ then $RST= 96 \times \frac{13}{4} = 312$, also not permitted.
In Case 1.4, we must use the digits $2,4,5,6$ and $7$ to construct the fraction $\frac{2}{13}$. Here $Q$ must be even, Also, we divide $PQ$ by $2$ and multiply by $13$, the last digit must be in the set of remaining digits above.
If $Q=2$, the possibilities for the last digit of $\frac{PQ}{2}$ are $1$ and $6$ which makes the possibilities for $T$ as being $3$ and $8$, neither of which are permitted. Hence, $Q=4$, in which case $T=6$ or $Q=6$, in which case, $T=4$. If $Q=4$ then $P$ must be even to give the right value for $T$ and so $P=2$ but $24 \times \frac{13}{2} = 156$, not permitted. If $Q=6$ then $P$ must be odd to give the right value for $T$ and we find $56 \times \frac{13}{2} = 364$ and $76 \times \frac{13}{2} = 494$, neither of which work for $RST$.

The only remaining case to analyse is $CD=14$.
If $CD = 14$ then $98 = 7 \times 14 > AB > 6 \frac{1}{2} \times 15 = 91$.
This leaves $5$ possibilities for $AB: 92, 93, 95, 96, 97$.
For each of the fractions the results of $7 - \frac{AB}{14}$ are $\frac{6}{14}, \frac{5}{14}, \frac{3}{14}, \frac{1}{7}$ an $\frac{1}{14}$, respectively, which we'll denote as cases $2.1, 2.2, 2.3, 2.4$ and $2.5$.
In Case 2.1, we must use the digits $3, 5, 6, 7$ and $8$ to construct the fraction $\frac{6}{14}$. Since, $PQ < 87$ here, it follows that $PQ \times \frac{13}{6} < 200$ which is not permitted for $RST$ since $R$ cannot be $1$.
In Case 2.2, we must use the digits $2,5,6,7$ and $8$ to construct the fraction $\frac{5}{14}$. Since we must then have $Q=5$, it follows that $PQ \leq 85 \times \frac{14}{5} = 238$ but this is also not permitted as if $R=2$ then $S$ must be greater than $4$ and $R$ cannot be less than $2$.
In Case 2.3, we must use the digits $2,3,6,7$ and $8$ to construct the fraction $\frac{3}{14}$. This means $PQ$ is a $2$-digit number, divisible by $3$ and hence, the sum $P+Q$ must also be divisible by $3$. This leaves the possibilities for $PQ$ as being $27, 36, 63$ and $72$. In each case, the results of $PQ \times \frac{14}{3}$ are $126$, $168$, $294$ and $336$, none of which work for $RST$.
In Case 2.4, we must use the digits $2,3,5,7$ and $8$ to construct the fraction $\frac{1}{7}$. Looking at residues, it must be the case that $Q \times 7 \equiv T$ mod $10$ with both being in the set above and being different. However, this is not the case for any two elements in the set. Hence, there is not solution in this case.
In Case 2.5, we must use the digits $2,3,5,6$ and $8$ to construct the fraction $\frac{1}{14}$. Looking again at residues, it must be the case that $Q \times 4 \equiv T$ mod $10$ with both being in the set above and being different. Hence, the only possibilities are $Q=3$ and $Q=8$ which, in both cases forces $T=2$. Checking through the possibilites for $RST$, we have
$53 \times 14 = 742$,
$63 \times 14 = 882$,
$83 \times 14 = 1162$,
$38 \times 14 = 532$,
$58 \times 14 = 812$,
$68 \times 14 = 952$,
none of which are appropriate for $RST$. Hence there are no solutions in this case.

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  • $\begingroup$ Some of the steps could have been more concise..but you are thorough ..detailed reasoning led to the right result $\endgroup$ – Uvc Jun 11 at 14:26
1
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One solution:

$7 = \frac{86}{13} + \frac{95}{247}$

How?

Let's first assume the following: $x \cdot CD = RST$, where it's likely that $x$ is an integer. Furthermore, $\frac{PQ}{RST}<1$, which means that $\frac{AB}{CD}>6$.

We have that

$AB > 6 \cdot CD$, and $AB < 7 \cdot CD$, which means that $CD$ is either $12, 13$ or $14$.


Upon editing this, and reviewing hexomino's answer, that our reasoning is pretty much identical, although not in the same order. I don't have time to write a complete explanation, especially when it's already written.

Therefore,

$A = 6, B=6, C=1, D=3, E=9, F=5, G=2, H=4, I=7$

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  • $\begingroup$ Your initial path is elegant..as hex completed this earlier, I have to give him the green check $\endgroup$ – Uvc Jun 11 at 14:25
  • $\begingroup$ $AB*RST+PQ*CD = 7CD*RST\implies AB*RST$ is divisible by $CD$. However, $6< \frac{AB}{CD}<7\implies RST$ is divisible by $CD$. So we always get that $RST=x CD$. $\endgroup$ – Geethu Joseph Jun 12 at 4:16

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