Question 894462
1. A grandfather and grandson celebrate their birthdays on the same day of the year, and for 6 consecutive years the grandfather's age is a multiple of the grandson's age.
SOLUTION:
on the first of those 6 consecutive years the grandfather is 61, and the grandson is 1,
on the second year, their ages are 62 and 2;
on the third year, their ages are 63 and 3;
on the fourth year, their ages are 64 and 4;
on the fifth year, their ages are 65 and 5, and
on the sixth year, the grandfather is 66, and the grandson is 6.
REASONING:
Starting with a very young grandson makes sense, so we start with a 1 year old grandson, who will be 2, 3, 4, 5, and 6 in the next 5 consecutive years.
If the grandfather's age on the first year is {{{x}}} ,
it is a multiple of the grandson's age, which is {{{1}}} ,
but in the next 5 years it must be that
{{{x+1}}} is a multiple of 2;
{{{x+2}}} is a multiple of 3;
{{{x+3}}} is a multiple of 4;
{{{x+4}}} is a multiple of 5, and
{{{x+5}}} is a multiple of 6.
That weans that {{{x-1}}} is a multiple of 2, 3, 4, 5, and 6.
The least common multiple of those numbers is {{{60}}} .
{{{x-1=60}}} ---> {{{x=61}}} gives the grandfather a reasonable age.
COMMENT:
Other common multiples of 2, 3, 4, 5, and 6, like 120, 180, and so on, make the grandfather way too old.
As another unrealistic option, we could start with a grandson who is 2 and a grandfather whose age minus 2 is a multiple of 2, 3, 4, 5, 6 and 7. Since the least common multiple is 210, that would make the grandfather at least 212. Making the grandson even older, makes the grandfather even more ridiculously old.
 
2. {{{sqrt(3)+7}}} is irrational, because {{{sqrt(3)}}} is irrational.
If {{{sqrt(3)+7}}} were rational, {{{sqrt(3)+7+(-7)=sqrt(3)}}} , being the sum of two rational numbers would be rational.
The fact that {{{sqrt(3)}}} is irrational is proven like the fact that {{{sqrt(2)}}} is irrational.
If {{{sqrt(3)}}} were rational, it could be written as an irreducible fraction,
{{{sqrt(3)=m/n}}} with {{{m}}} and {{{n}}} not having any common factors.
Then,
{{{sqrt(3)=m/n}}} ---> {{{(sqrt(3))^2=(m/n)^2}}} ---> {{{3=m^2/n^2}}} ---> {{{3n^2=m^2}}} .
That means that {{{m^2}}} is a multiple of 3,
but then {{{m}}} must be a multiple of 3,
which would make {{{m^2}}} a multiple of 9, and {{{m^2/3=n^2}}} would be a multiple of 3, just like {{{m}}} is.
That contradicts the starting hypothesis that {{{sqrt(3)=m/n}}} with {{{m}}} and {{{n}}} not having any common factors.
So, it is impossible to find a rational number {{{m/n}}} with {{{m}}} and {{{n}}} not having any common factors, and {{{(m/n)^2=3}}} .
In other words, it is impossible to find a rational number equal to {{{sqrt(3)}}} , so {{{sqrt(3)}}} is irrational.