Simple problem but how to solve

[Guest]

1) find the last two-digits of
15*37*63*51*97*17
options are:
(A)35 (B)45 (C)55 (D)85 .
ans. is 35 how please explain?

2)what will be the remainder when 19100 is divided by 20?
(A)19 (B)20 (C)3 (D)1 .
ans is 1 how??

[Guest]

1. You should find [tex]15*37*63*51*97*17 (mod 100)[/tex].
[tex]15*37*63*51*97*17 \equiv 555*3213*1649 \equiv 55*13*49 \equiv 715*49 \equiv 15*49 \equiv 735 \equiv 35 (mod 100)[/tex]

2. 19111 is divisible by 20. Therefore the remainder is equal to 0.

[Guest]

The previous post is right the answers are 35 and 0.

An easier way to calculate question 1, is to use the fact that the chinese remainder theorem says that to calculate something mod 100 it is enough to calculate it mod 4 and mod 25 (as 4*25=100 and 4 and 25 share no common factors other than 1).

Calculating the product mod 4 is very easy:
15*37*63*51*97*17 = 3*1*3*3*1*1 = 3 mod 4

Calculating the product mod 25 is a bit trickier. Notice that we have 15 in the product so it must be a multiple of 5 (a factor of 25).

Calculating (product/5) mod (25/5) is very easy:
3*37*63*51*97*17 = 3*2*3*1*2*2 = 2 mod 5
This implies product/5=5n+2 for some integer n, so product=25n+10, therefore the product mod 25 = 10.

So product=3 mod 4 and product=10 mod 25, we could use the chinese remainder theorem to calculate product mod 100 but it is easier to just realise that product=10 mod 25 implies
product = 10, 35, 60, or 85 mod 100
and only 35 is 3 mod 4.

So the answer is 35.

Hope this helped,

R. Baber.

[Guest]

I checked online, and found that the second question was incorrectly transcribed, it should be:
what will be the remainder when [tex]19^{100}[/tex] is divided by [tex]20[/tex]?

[tex]19^{100} = (-1)^{100} = 1\; \bmod\; 20[/tex]

Basically you are using the fact that
[tex](xn+a)(yn+b) = xyn^2+bxn+ayn+ab = (xyn+bx+ay)n+ab[/tex]
So [tex](xn+a)(yn+b) \bmod n[/tex] is [tex]ab \bmod n[/tex],
or put another way (let [tex]c=xn+a[/tex], and [tex]d = yn+b[/tex])
[tex]cd \bmod n[/tex] is [tex](c \bmod n)(d \bmod n) \bmod n[/tex].
which generalises to
[tex]c_1c_2\cdots c_m \bmod n[/tex] is [tex](c_1 \bmod n)(c_2 \bmod n)\cdots(c_m\bmod n) \bmod n[/tex].
If [tex]c_1=c_2=\cdots=c_m[/tex] then this tells us that
[tex]c_1^m \bmod n[/tex] is [tex](c_1 \bmod n)^m \bmod n[/tex].

[tex]19^{100} \bmod 20 =(19 \bmod 20)^{100}\bmod 20[/tex]
[tex]=(-1 \bmod 20)^{100} \bmod 20[/tex]
[tex]=(-1)^{100} \bmod 20[/tex]
[tex]=1 \bmod 20[/tex]

Hope this helped,

R. Baber.

[Guest]

Actually it might just be easier to look at (product/5) mod (100/5) for question 1:
3*37*63*51*97*17 = 3*-3*3*-9*-3*-3 = 27*9*9 = 27*81 = 7 mod 20
So product = 7*5 = 35 mod 100.

R. Baber.

[Guest]

Since [tex]a_{n+1}=F(a_n)[/tex], if the sequence is convergent with limit [tex]l[/tex], then this limit fulfills the equation [tex]l=F(l)[/tex]. The roots of this equation are 4 and 6.
First of all we will examine the values of [tex]a_n[/tex] for which [tex]a_{n+1}\ge a_n[/tex].
This means solving the inequality [tex]F(x)>x[/tex]. The idea is to use an important theorem for convergent sequences - a sequence which is monotone and bounded is convergent. The inequality holds for [tex]x \in (-12,4) \cup (6,+\infty)[/tex] respectively for [tex]a_n[/tex] in these intervals (*).
We will examine for what initial values [tex]a_1[/tex], [tex]a_2[/tex] lies in some of the 4 intervals (these intervals in which the real line is divided by the solution of the inequality).
1) [tex]a_2>6[/tex], so [tex]F(a_1)>6[/tex] and we find [tex]a_1 \in (-12,-4)\cup (6,+\infty)[/tex]. Then [tex]a_3=F(a_2)>a_2>6[/tex] and by induction [tex]a_{n+1}=F(a_n)>a_n>6[/tex] (we know that [tex]F(x)>x[/tex] for [tex]x>6[/tex]). Thus for these initial values the sequence is monotone. However it is not bounded - otherwise it would have limit which however is strictly greater then 6 - impossible as we saw in the beginning. Hence the sequence diverges.
2)[tex]6>a_2>4[/tex], so [tex]6>F(a_1)>4[/tex] and [tex]a_1\in (-4,-3)\cup (4,6)[/tex]. Thus [tex]a_n \in (4,6)[/tex] (since we see from the solutions that when [tex]x\in(4,6)[/tex] then [tex]F(x)\in(4,6)[/tex]). So for such initial value the sequence is bounded in [tex](4,6)[/tex] (after the first term). From (*) it follows that the sequence is decreasing. Hence its limit is 4.
3) [tex]4>a_2>-12[/tex], so [tex]4>F(a_1)>-12[/tex] and [tex]a_1\in (-3,4)[/tex]. In this case [tex]F(a_2)>a_2[/tex]. We should examine precisely where in [tex](-12,4)[/tex] [tex]a_2[/tex] lies in order to see where goes [tex]F(a_2)[/tex]. Observe that [tex]F(x)=a[/tex] has solution for [tex]a\in (-\inft,-94)\cup (2,+\infty)[/tex]. Thus [tex]a_2 \in (2,4)[/tex]. For such value we know that [tex]F(a_2)>a_2[/tex]. However [tex]F(a_2)<4[/tex] since [tex]a_2\ in (-3,4)[/tex](in the beginning of that point) ([tex]F(x)>x[/tex] for [tex]x\in (2,4)[/tex] and [tex]F(x)<4[/tex] in that interval). Inductively we proof that [tex]a_n<4[/tex] and is increasing, so the limit is 4.
4) [tex]a_2<-12[/tex], so [tex]F(a_1)<-12[/tex] and [tex]a_1\in (-\infty,-12)[/tex]. This means that [tex]a_n<-12[/tex] for all [tex]n[/tex] so the sequence cannot converge (the limit point candidates are 4 and 6). Actually it diverges.

Now it remains to write the answer
For [tex]a_1 \in (-12,-4)\cup (6,+\infty)[/tex] - divergent
For [tex]a_1\in (-4,-3)\cup (4,6)[/tex] - convergent with limit 4
For [tex]a_1\in (-3,4)[/tex] - convergent with limit 4
For [tex]a_1\in (-\infty,-12)[/tex] - divergent
For [tex]a_1=-3[/tex] and [tex]a_1=4[/tex] - convergent with limit 4
For [tex]a_1=-4[/tex] and [tex]a_1=6[/tex] - convergent with limit 6

[leesajohnson]

If you will multiply these numbers you will get 2940521535 so the last digit is 35.