ALPHAMATICS

Level pending

The following Alphamatic has different letters that stand for different digits 0, 1, 2, 3...9 :

F O R T Y + T E N + T E N = S I X T Y \text{F O R T Y + T E N + T E N = S I X T Y}

Find the value of S + I + X + T + E + E + N \text{ S + I + X + T + E + E + N} .


The answer is 26.

Sanyam Khankhoje by sanyam khankhoje , 21, India

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1 solution

Pranjit Handique
Feb 27, 2014

Observing the digits in the units place N is either 0 or 5. if N=5 there will be carryover of 1 to the Ten's place. Again by oberving the digits in the Ten's position 2E+1 must have a 0 in the unit palce which is not possible in this case. So N=0 and E=5. So the carryover into the Hundred's place is 1. We can see there is a definite carryover to the Thousand's and TenThousand's place. The maximum carryover to the Thousand's place can be 2.(It can be 1 or 2). If the carryover is 1 then O has to be 9( less than 9 wont produce a carryover to the TenThousand's place) to produce a carryover to the TenThousand's place and I has to be 0 but which is impossible(0 has been already assigned to N). So the carryover to the Thousand's palce is 2 and for the similar reasons O=9 so that I=1(if O=8 then again I will be 0) and the carryover to the TenThousand's place is 1. Thus S=F+1.

We are left with 2,3,4 and 6,7,8.Let us concentrate on the digits in the Hundred's place. T cannot be 2,3,4 because 2T+R (even with the highest value for T=4 and R=8 does not produce a carryover of 2 into the TenThousand's place. There are three possible value for R and T i.e 6,7,8. T=6 and R=7 is not possible because in that case X will become 0 (0 has been already assigned to N). Again T=6 and R=8 is not possible beacuse in that case x becomes 1( 1 has been already assigned to I). So T is definitely not 6. T is either 7 or 8. R=6 and T=7 is not possible beacause X agains becomes 1. if R=6 and T=8 X becomes 3 which will leave the values 2,4,6,8 left for F and S which will not satisfy S=F+1 as noticed earlier. So R is definitely not 6. R and T must be either 7 or 8. if R=8 and T=7 then again X=3 whih is again impossible because it does not satisfies S=F+1. So, definitely R=7 and T=8. Thus X=4 leaving the only values 2 and 3 which satisfies S=F+1. So, F=2 and S=3. Only digit remaining is 6. Y=6.

So, N=0, I=1, F=2, S=3, X=4, E=5, Y=6, R=7, T=8, O=9,

SIXTEEN = 3+1+4+8+5+5+0=26.

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Observing the digits in the units place N is either 0 or 5. if N=5 there will be carryover of 1 to the Ten's place. Again by oberving the digits in the Ten's position 2E+1 must have a 0 in the unit palce which is not possible in this case. So N=0 and E=5. So the carryover into the Hundred's place is 1. We can see there is a definite carryover to the Thousand's and TenThousand's place. The maximum carryover to the Thousand's place can be 2.(It can be 1 or 2). If the carryover is 1 then O has to be 9( less than 9 wont produce a carryover to the TenThousand's place) to produce a carryover to the TenThousand's place and I has to be 0 but which is impossible(0 has been already assigned to N). So the carryover to the Thousand's palce is 2 and for the similar reasons O=9 so that I=1(if O=8 then again I will be 0) and the carryover to the TenThousand's place is 1. Thus S=F+1.

We are left with 2,3,4 and 6,7,8.Let us concentrate on the digits in the Hundred's place. T cannot be 2,3,4 because 2T+R (even with the highest value for T=4 and R=8 does not produce a carryover of 2 into the TenThousand's place. There are three possible value for R and T i.e 6,7,8. T=6 and R=7 is not possible because in that case X will become 0 (0 has been already assigned to N). Again T=6 and R=8 is not possible beacuse in that case x becomes 1( 1 has been already assigned to I). So T is definitely not 6. T is either 7 or 8. R=6 and T=7 is not possible beacause X agains becomes 1. if R=6 and T=8 X becomes 3 which will leave the values 2,4,6,8 left for F and S which will not satisfy S=F+1 as noticed earlier. So R is definitely not 6. R and T must be either 7 or 8. if R=8 and T=7 then again X=3 whih is again impossible because it does not satisfies S=F+1. So, definitely R=7 and T=8. Thus X=4 leaving the only values 2 and 3 which satisfies S=F+1. So, F=2 and S=3. Only digit remaining is 6. Y=6.

So, N=0, I=1, F=2, S=3, X=4, E=5, Y=6, R=7, T=8, O=9,

SIXTEEN = 3+1+4+8+5+5+0=26.

Piyush Patnaik - 7 years, 3 months ago

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