Em(p)ty

Logic Level 1

In the following equation, each letter represents a distinct digit in base ten:

Y E × M E T T T \large {\begin{array}{cccccc} & & &&& Y&E\\ \times& & & & & M&E\\ \hline & & & & T & T&T\\ \end{array}}

Determine the sum E + M + T + Y . E +M +T +Y.


The answer is 21.

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Watsapol Kerdlarppol by Watsapol Kerdlarppol , 22, Thailand

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1 solution

Anirudh Sreekumar
Mar 29, 2017

( Y E ) ‾ ⋅ ( M E ) ‾ = T T T ‾ ( Y E ) ‾ ⋅ ( M E ) ‾ = T ⋅ 111 = T ⋅ 37 ⋅ 3 ( 1 ) \begin{aligned}\overline{(YE)}\cdot\overline{(ME)}&=\overline{TTT}\\ \overline{(YE)}\cdot\overline{(ME)}&=T\cdot 111\\ &=T\cdot37\cdot3&\small\color{#3D99F6}(1)\end{aligned}

One of ( Y E ) ‾ , ( M E ) ‾ \overline{(YE)},\overline{(ME)} has to be 37 37 or 74 74

Case 1: Let, Y E ‾ = 74 \color{#3D99F6}\text{Case 1: }\color{#333333}\text{Let,}\overline{YE}=74

This is only possible if T = 8 Else, M E ‾ < 10 The last digit of ( Y E ) ‾ ⋅ ( M E ) ‾ is given by, ( 10 Y + E ) ( 10 M + E ) ( m o d 10 ) ≡ T T T ‾ ( m o d 10 ) ⟹ E 2 ( m o d 10 ) ≡ T However, E 2 ( m o d 10 ) ≡ 1 , 4 , 5 , 6 , 9 , 0 Perfect squares always end in (1,4,5,6,9,0) Thus, T ≠ 8 Thus YE cannot be equal to 74 \begin{aligned}\text{This is only possible if} \hspace{5mm}T=8&\hspace{9mm}\small\color{#3D99F6}\text{Else, }\overline {ME}<10\\\text{The last digit of} \overline{(YE)}\cdot\overline{(ME)}\text {is given by,}\\ (10Y+E)(10M+E)\pmod{10}&\equiv\overline {TTT} \pmod{10}\\ \implies E^2\pmod{10}&\equiv T\\ \text{However,} E^2\pmod{10}&\equiv1,4,5,6,9,0& \color{#3D99F6}\small\text{Perfect squares always end in (1,4,5,6,9,0)}\\ \text{Thus,} T&\neq8 \\ \text{Thus YE cannot be equal to 74}\\&\\\end{aligned}

Case 2: Let, Y E ‾ = 37 ⟹ E = 7 ⟹ ( M E ) ‾ = 3 T From (1) 3 T ( m o d 10 ) = 7 ⟹ T = 9 ( M E ) ‾ = 3 T = 27 Thus we have, Y = 3 , M = 2 , E = 7 , T = 9 E + M + T + Y = 21 \begin{aligned}\color{#3D99F6}\text{Case 2: }\color{#333333}\text{Let,}\overline{YE}=37\\ \implies E&=7\\ \implies\overline{(ME)}&=3T\hspace{8mm}\small\color{#3D99F6}\text{From (1)}\\ 3T\pmod{10}&=7\\ \implies T&=9\\ \overline{(ME)}&=3T=27\\ \text{Thus we have,}\\ &\color{#D61F06}\boxed{Y=3},\boxed{M=2},\boxed{E=7},\boxed{T=9}\\ &\color{#D61F06}E+M+T+Y=\boxed{21}\end{aligned}

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I disagree with "One of them has to be 37" Why can't one of them be 74?

Calvin Lin Staff - 4 years, 2 months ago

@Calvin Lin - oops i missed the case of E=8 ,i'll fix it now

Anirudh Sreekumar - 4 years, 2 months ago

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Can you clean up the presentation of the solution? Right now, you know what you want to show, and you're thinking it, but you're not showing it to the reader.

IE It would be clearer to say "One of YE, ME has to be 37 or 74. Case 1: equal 37 ... Case 2: equal 74 ..." Otherwise, you just have a bunch of mathematical equations, without explaining why they are written down.

Calvin Lin Staff - 4 years, 2 months ago

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