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Задачи с Наименьшим Общим Кратным (НОК)
В этом уроке мы рассмотрим некоторые текстовые задачи с Наименьшим Общим
Кратным (НОК). Вы можете встретить эту величину либо в чистом виде – у
Вас попросят вычислить НОК для двух или более чисел – либо поиск НОК
станет необходимостью при решении текстовой задачи. Именно о последнем
варианте развития событий мы и поговорим сегодня.
In this lesson we consider some Math Word problems with the Lowest Common Multiple (LCM). We have two chances to meet it: the problem can ask simply about value of the LCM of two or more integers, or we can face the finding the common quantity for some word-expressed problem. Let’s look at the second variant .
There are two simple algorithms of finding the LCM.
If we have two positive integers a and b, then we can write them as products of primes:
a=p*q*r,
b=p*r*s*t, where p,q,r,s,t are primes.
Then LCM(a,b)=p*r*q*s*t.
Or, in the other words, the LCM is a product of all primes, that are divisors of a and b, and those of them, which are common for a and b, we write only once.
If we have two positive integers a and b, then we can write them as products of powers of primes, for example:
a=x^2 y^3 z^6,
b=x^5 y^1, where x, y, z are primes.
Then LCM(a, b)= x^5 y^3 z^6.
Or, in the other words, the LCM is a product of the maximum powers of all primes, that are divisors of a and b.
There are some examples of such problems and correspondent solutions.
The traffic lights at three different road crossings change after every 48s, 72s, and 108s respectively. If they all change simultaneously at 08:20:00 A.M., then at what time will they again change simultaneously?
A 08:22:24
B 08:23:36
C 08:27:12
D 08:29:06
E 08:31:44
To answer this question we need to find the minimum number of seconds that will be evenly divisible by 48, 72 and 108. It will be the LCM(48, 72, 108). Let’s write down prime factorizations for all these numbers:
48 = 16*3 =2^4 3^1,
72 =8*9 = 2^3 3^2,
108 =4*27 =2^2 3^3,
So, LCM(48, 72, 108) =2^4 3^3 =432. Here we used the second rule of finding the LCM.
432 seconds = 7 minutes 12 seconds, so, the answer is C.
At a certain race, two athletes run around a stadium. The first one runs a lap in 6 minutes; the second one - in 10 minutes. If they start running simultaneously, then how long will it take the first athlete to be exactly two laps ahead of the second one?
A 20 min.
B 25 min.
C 30 min.
D 35 min.
E 40 min.
To answer this question we need to find the minimum number of minutes that will be evenly divisible by 6 and 10. It will be the LCM(6, 10). Let’s write down prime factorizations for these numbers:
6 =2*3,
10=2*5,
So, LCM(6, 10) =2*3*5 =30 minutes. Here we used the first rule for finding the LCM.
In 30 minutes the first athlete will run 30/6 = 5 laps and the second 30/10 =3 laps, so, the first athlete will be exactly two laps ahead of the second one. The answer is C.
Resume. This lesson covers types of problems, that face calculating of the LCM and two rules, that help us to count it in a right way . You should know right and fast solution, like the solutions described in the lesson, in order to finish the GMAT on time.
Material prepared by Ksenia Zueva, GMAT consultant at MBA Strategy
