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4 years ago

In 1998, Cathy's age is equal to the sum of the four digits in the year of her birthday, then how old was Cathy in 1998?

Mathematics

1 answer:

ycow [4]4 years ago

5 0

Answer:

<em>Cathy was born in 1980 and she was 18 years old in 1998</em>

Step-by-step explanation:

<u>Equations</u>

This is a special type of equations where all the unknowns must be integers and limited to a range [0,9] because they are the digits of a number.

Let's say Cathy was born in the year x formed by the ordered digits abcd. A number expressed by its digits can be calculated as

$x=1000a+100b+10c+d$

In 1998, Cathy's age was

$1998-(1000a+100b+10c+d)$

And it must be equal to the sum of the four digits

$1998-(1000a+100b+10c+d)=a+b+c+d$

Rearranging

$1998=1001a+101b+11c+2d$

We are sure a=1, b=9 because Cathy's age is limited to having been born in the same century and millennium. Thus

$1998=1001+909+11c+2d$

Operating

$88=11c+2d$

If now we try some values for c we notice there is only one possible valid combination, since c and d must be integers in the range [0,9]

c=8, d=0

Thus, Cathy was born in 1980 and she was 18 years old in 1998. Note that 1+9+8+0=18

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