Logical Numbers

Logic Level 3

All digits 1 to 9 are used in place of the letters A through I such that

  • A + B + C A+B+C > D + E + F D+E+F > G + H + I G+H+I

  • E E is a prime factor of G G

  • F F > A A

  • B + G = H B+G= H

  • I I is not 1 1

Which letter represents which number?

Concatenate the answer as the nine-digit integer A B C D E F G H I \overline{ABCDEFGHI} .


The answer is 719628453.

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4 solutions

Satyen Nabar
Oct 13, 2015

E must be either 2/3.

G must be either 4/6/8/9.

Since sum of all numbers is 45, the maximum that G+H+I can be is 14.

If I=2, then E=3 and G=6. Which means H is greater than 6. And then G+H+I >14 so incorrect.

If I=3, E=2, G= 4/6/8. Since G+H+I is the least G cant be 6/8. G=4.

H can be be max 7, so B can only be 1/2/3 but 2/3 are used up. So B=1. H=5.

ACDF can be 6/7/8/9 in some order. Since A+B+C>D+E+F and F>A, C must be at least 3 greater than D. Thus C=9, D= 6, F=8, A=7.

719628453

Why ans is not 123754689???

Durgesh Kulkarni - 5 years, 8 months ago

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Because A+B+C > D+E+F > G+H+I.

Satyen Nabar - 5 years, 8 months ago

bcoz E is prime factor of G

kaushal dhaundiyal - 5 years, 7 months ago

Why is it not 836529473 ?

Sidhartha Khare - 5 years, 7 months ago

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All the alphabets represent different numbers but you have used 3 two times.

Praful Jain - 5 years, 7 months ago

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I think there are many solutions. 3 is also a prime no. 125 439 687 also 721 439 685 not changing the position of B, E, G & H after solving once, I have found many solutions like that. Please comment on that.

Anirban Bhattacharjee - 5 years, 7 months ago

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@Anirban Bhattacharjee It should be A+B+C>D+E+F>G+H+I

Praful Jain - 5 years, 7 months ago

Because you have used 3 twice. And not used 1.

Satyen Nabar - 5 years, 7 months ago
Saya Suka
Mar 23, 2021

G must be big enough to have at least a prime factor other than itself but small enough so that the sum of itself and two other digits — at least one of those larger than G itself — is less than 15.

After considering that I ≠ 1, G can only be 4, E = 2, I = 3, B = 1 and H = 5.

A + 1 + C > D + 2 + F
&
F > A
==> C = 9


A + B + C + D + E + F
= 9(9 + 1)/2 - (G + H + I)
= 45 - (4 + 5 + 3)
= 33
A + B + C > 33/2 = 16.5
A + B + C = 17
A = 17 - 1 - 9 = 7 ==> A = 7
==> F = 8
==> D = 6

Answer = 719628453

John Williamson
Sep 18, 2016

I wrote up this explanation for a friend in India. Anyone is welcome to use it. I changed "I" to "J" so it was obviously different from "1".

Viet Ký
Oct 23, 2015

Here is my answer (copy it to chrome console): my link to answer

The code is not optimized but produces the right answer.

Sorry for cheating :)

This question should be solved by thinking as an easier way.

Lu Chee Ket - 5 years, 4 months ago

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