MHB Man's Age Now: Find Out His Digit Sum Born Year

  • Thread starter Thread starter Albert1
  • Start date Start date
Albert1
Messages
1,221
Reaction score
0
A man found (at year 2011) his age at that time was all the digit sum of the year when he was born,
can you tell me ,how old is he now ?
 
Mathematics news on Phys.org
Albert said:
A man found (at year 2011) his age at that time was all the digit sum of the year when he was born,
can you tell me ,how old is he now ?
[sp]The man was born in the year 1991 so that in the year 2011 his age was 1 + 9 + 9 + 1 = 20 years ...[/sp]

Kind regards

$\chi$ $\sigma$
 
the sum of digits cannot be > 28 as 3 digit number < 2011 largest sum of digits (1999) is 28
so the year is 19ab or 20cd ( digit positions)

take 19ab.
now 10 a+ b + 1+ 9 + a + b = 111
or 11a + 2b = 111 so a is odd
a = 9 , b = 1 giving 1991
a = 7 or below give b > 9 not possible

take 20cd
so 10c + d + 2 + c + d = 11
11c + 2 d = 9
c = 0 and d = fraction
so no solution
so only solution 1991 and age = 20
 
I mean how old is he now ?(year 2015)
 
Albert said:
I mean how old is he now ?(year 2015)

24 years now
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Back
Top