Question

Determine which digits do the letters represent.
ABCD + EFGB = EFCBH​

Answer 1

Answer:

Hey mate here is your answer

Step-by-step explanation:

1234 + 5672 = 56328

But correct calculation will be 6906

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Answer 2

Answer:

The digits of the given letters are given below

$9 & 5 & 6 & 7 & 1 \\ & 1 & 0 & 8 & 5 \\\\1 & 0 & 6 & 5 & 2 &$

Step-by-step explanation:

• Since the addition of two digits will give a maximum of 18 , or a maximum of 19 if there is a carryover (or renaming) of 1 , this means that the maximum carryover is $1 .$
• So $E=1 .$
• In the thousands column, $A+E=10+F$ or $A+E+1=10+F$, and since $E =1$, then$F=A-9 \ or\ A-8 .$
• Now $A \leq 9 , so F=0 or 1$ . But $E=1$ already, so $F=0$.
• In the hundreds column, if $B+F=10+C ,$ then $B=10+C$ is not possible. If $B+F+1=10+C$, then $B=9+C$ means $B=9$ and $C=0$, but $F=0$ already.
• This means that there is no carryover from $B+F$ or $B+F+1 .$
• But$B+F \neq C$ as $F=0$ and $B \neq C$, so $B+F+1=C$, i.e. there is carryover from the tens column. Since $F=0$, then
• $\mathbf{B}+\mathbf{1}=\mathbf{C} . - - (1)$
• This also means that in the thousands column, $A+E=10 ,$ i.e. $\mathbf{A}=\mathbf{9}$ since$E =1$.
• suppose there Is no carryover in the ones column, i.e. $D+b=H$.
• Then $\mathrm{C}+\mathrm{G}=10+\mathrm{B}$ because there is carryover from the tens column.
• From (1), $C=B+1$, so $B+1+G=10+B$, i.e. $G=9$. But $A=9$ already.
• So there is a carryover in the ones column, i.e. $D+B=10+H_{.}---(2)$
• Then $C+G+1=10+B$, i.e.$B+1+G+1=10+B ,$ which gives $\mathbf{G}=\mathbf{8}$.
• From $(2), \mathrm{D}+\mathrm{B}=10+\mathrm{H} .$
• Since $D, B \leq 7 (as G=8$ and $A=9$ already) and $D \neq B$, the maximum value of $D+B$ is $7+6=13$, i.e. $H \leq 3$.
• If $B=7$, from (1), $C=8$, but $G=8$ already.
• If $B=6$, then $D=7$. But from $(1), C=7$ also, a contradiction. So $\mathrm{H} \neq 3 .$
• But $\mathrm{H} \geq 2$ (as $F=0$ and $E=1$ already), so $\mathbf{H}=\mathbf{2}$. Then $D+B=7+5$ or $5+ 7 .$
• If $B=7$, from $(1), C=8$, but $G=8$ already.
• So$\mathbf{D}=\mathbf{7}$and $\mathbf{B}=\mathbf{5}$. From (1), $\mathbf{C}=\mathbf{6} .$
• Thus $E=1, F=0, A=9, G=8, H=2, D=7, B=5, C=6$ :
• $9 & 5 & 6 & 7 & 1 \\ + & 1 & 0 & 8 & 5 \\\\1 & 0 & 6 & 5 & 2 &$