Number Spy Challenge 11

Find a number that when divided by the Hardy-Ramanujan taxicab number 11, it produces a number such that the sum of it's digits is equivalent to a quarter of a power of 22.

In simplified terms:

x1729\frac{x}{1729} =abcd= abcd such that a+b+c+d=a + b + c + d = 2n4\frac{2^n}{4}

The only condition is that x1729x \geq 1729 as 17291729\frac{1729}{1729} =1= 1.

You're looking for the first number.

Note by Yajat Shamji
4 months, 3 weeks ago

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1 vote

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Comments

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@Zakir Husain, @Pi Han Goh, @Vinayak Srivastava, @Mahdi Raza, @Aryan Sanghi

Yajat Shamji - 4 months, 3 weeks ago

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@Yajat Shamji- Is abcd=a×b×c×dabcd=a\times b\times c \times d or they are digits?

Zakir Husain - 4 months, 3 weeks ago

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I think abcdabcd is the four-digit number. Yes, I agree that it might be confusing because the result can be more than a four-digit number, but let him clarify

Mahdi Raza - 4 months, 3 weeks ago

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@Gandoff Tan

Yajat Shamji - 4 months, 3 weeks ago

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As xx can be equal to 17291729 (x1729\because x\geq1729). If x=1729x=1729 then a=0,b=0,c=0,d=1;a+b+c+d=1=224a=0,b=0,c=0,d=1;a+b+c+d=1=\dfrac{2^2}{4} therefore x=1729x=1729

Zakir Husain - 4 months, 3 weeks ago

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17291729=0001=0+0+0+1=1=224\dfrac{1729}{1729} = 0001 = 0 + 0 + 0 + 1 = 1 = \dfrac{2^2}{4}

There might be infinite of these

Mahdi Raza - 4 months, 3 weeks ago

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It is written that x1729=abcd;a+b+c+d=2n4\dfrac{x}{1729}=abcd;a+b+c+d=\dfrac{2^n}{4} not x1729=2n4\red{\dfrac{x}{1729}=\dfrac{2^n}{4}}

Zakir Husain - 4 months, 3 weeks ago

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What is asked is the sum of digits after the division and that is 1

Mahdi Raza - 4 months, 3 weeks ago

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@Mahdi Raza thanks!

Zakir Husain - 4 months, 3 weeks ago

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@Mahdi Raza No. What is asked is that the sum of the digits equals a quarter of a power of 22.

Yajat Shamji - 4 months, 3 weeks ago

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Thanks for finding the first number, @Mahdi Raza, @Zakir Husain!

Yajat Shamji - 4 months, 3 weeks ago

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@Mahdi Raza, @Zakir Husain - I am posting Number Spy Challenge 22 in 5\leq 5 minutes! After that, I'll see you tomorrow!

Yajat Shamji - 4 months, 3 weeks ago

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