# Arithmetic: answer key to the exercises in the final year of high school in PDF.

Exercise of arithmetic in terminale S.
1-Position d = gcd(a,b)
We have if d divides a and d divides b then d divides b and d divides (a-bq)
Reciprocally: if d divides b and d divides (a-bq) then d divides ( a – bq ) +bq = a
2- it is the previous relationship with .

Demonstrate a property, exercise of mathematics in terminale S on arithmetic.

Let δ = PGCD(a ;b) and µ = PPCM(a ;b).
Then we have a= δa’ and b = δb’ with a’ and b’ prime between them.
So we have PPCM(a’ ;b’) = a’b’
µ = PPCM(δa’; δb’) = δ×PPCM(a’; b’) = δ×a’×b’
Thus δµ = δ²×a’×b’ = δ×a’× δ×b’ = ab

System of equations and arithmetic.

From the relation PPCM(a;b)×PGCD(a;b) = ab,
we deduce from the second equation of the system :
3×PPCM(a ;b) = PGCD(a ;b)×PPCM(a ;b)
So PGCD(a;b) = 3
Then there are integers a’ and b’ (prime to each other) such that a = 3a’ and b = 3b’.
Carrying over into the first equation, we get :
9a’² – 9b’² = 405
Let (a’ + b’)(a’ – b’) = 45
Now 45 = 3²×5
We can therefore have the following 3 systems:
a’ = 23 and b’ = 22 a’ = 9 and b’ = 6 a’ = 7 and b’ = 2
Hence (a ;b) = (66 ;69) is not suitable because hence (a ;b) = (42 ;6)
9 and 6 are not prime to each other
The solution pairs of the system are therefore (42 ;6) and (66 ;69).

Prime numbers .

1)
2)
because a and b are greater than or equal to 2.
is the product of two integers greater than 1.
So is not a prime number.

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## The exercises in the final year

After having consulted the answer key of this exercise arithmetic in senior year, you can return to the exercises in senior year

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