Math 25: Solutions to Homework # 4 (4.3 # 10) Find an integer that
... The total cost was x42y cents. If 88 divides this number, then both 8 and 11 must divide it. We know that 8 | x42y if 8 | 42y. Thus we must have y = 4 since 424 is the only number of this form divisible by 8. Now 11 | x424 only if 11 divides x − 4 + 2 − 4 = x − 6. Thus x = 6, so the total cost was ...
... The total cost was x42y cents. If 88 divides this number, then both 8 and 11 must divide it. We know that 8 | x42y if 8 | 42y. Thus we must have y = 4 since 424 is the only number of this form divisible by 8. Now 11 | x424 only if 11 divides x − 4 + 2 − 4 = x − 6. Thus x = 6, so the total cost was ...
Course of analytical geometry
... Like the elementary geometry explained in the book [6], the analytical geometry in this book is a geometry of threedimensional space E. We use the symbol E for to denote the space that we observe in our everyday life. Despite being seemingly simple, even the empty space E possesses a rich variety of ...
... Like the elementary geometry explained in the book [6], the analytical geometry in this book is a geometry of threedimensional space E. We use the symbol E for to denote the space that we observe in our everyday life. Despite being seemingly simple, even the empty space E possesses a rich variety of ...
Bounded length intervals containing two primes and an almost
... We wish to show, for a suitable choice of positive constants B and k, that S > 0 for any large N. If S > 0 for some N, then at least one term in the sum over n must have a strictly positive contribution. Since the λd are all reals, we see that if there is a positive contribution from n ∈ [N, 2N), th ...
... We wish to show, for a suitable choice of positive constants B and k, that S > 0 for any large N. If S > 0 for some N, then at least one term in the sum over n must have a strictly positive contribution. Since the λd are all reals, we see that if there is a positive contribution from n ∈ [N, 2N), th ...
