# Thread: A few problems with solutions..?

1. ## A few problems with solutions..?

What is the smallest number greater than 1 that, when divided by 2, 3, 4, 5 or 6 leaves a remainder of 1 in each case?

How do get 61 as the answer?

Elkville High won a Friday night b-ball game by 10 points; the next night they scored 25 points more than on Friday and again won by 10 points. The sum of the opponents' scores for the 2 games was 109. How many points did Elkville score on Friday?

Again, how do you get 52 as the answer?

If 537^102 were calculated, it would have 279 digits. What would the digit farthest to the right be (the ones digit)?

Why is the answer 9??

And lastly, what is the smallest possible value for the product of two real numbers that differ by 6?

The answer is -9 but why??

2. Originally Posted by sarahh
What is the smallest number greater than 1 that, when divided by 2, 3, 4, 5 or 6 leaves a remainder of 1 in each case?

How do get 61 as the answer?
the number we are looking for is of the form: $n + 1$, where $n$ is the LCM of 2,3,4,5, and 6. thus the answer is 60. do you see why we needed to find the LCM?

3. No I don't.. if you don't mind, could you provide a little more info?

4. Originally Posted by sarahh
...

Elkville High won a Friday night b-ball game by 10 points; the next night they scored 25 points more than on Friday and again won by 10 points. The sum of the opponents' scores for the 2 games was 109. How many points did Elkville score on Friday?

Again, how do you get 52 as the answer?

...
Hi,

make a table:
1. game:
scores of EH: x
scores of opponent: x-10

2. game:
scores of EH: x+25
scores of opponent: x+25-10

According to the problem

(x-10) + (x + 25 - 10) = 109
2x + 5 = 109
2x = 104
x = 52

5. Originally Posted by sarahh
...
If 537^102 were calculated, it would have 279 digits. What would the digit farthest to the right be (the ones digit)?

Why is the answer 9??

...
Hi,

Consider the following multiplications (? stands for any digit)

$(????7)^1 = ????7$

$(????7)^2 = ??...??9$.........because $7 \cdot 7 = 4\boxed{9}$

$(????7)^3 = ??...??3$.........because $7 \cdot 9 = 6\boxed{3}$

$(????7)^4 = ??...??1$.........because $7 \cdot 3 = 2\boxed{1}$

$(????7)^5 = ??...??7$.........because $7 \cdot 1 = \boxed{7}$

That means after 4 steps the pattern repeat again.

102 contains 25 groups of 4 (which will not change the last digit) and 2 single steps. Look at the result at the 2nd step: 9

6. Originally Posted by sarahh
...
And lastly, what is the smallest possible value for the product of two real numbers that differ by 6?

The answer is -9 but why??
Hi,

1. number: x
2. number: x-6

Theproduct of these 2 numbers is:

p(x) = x(x-6) = x² - 6x

You'll get an extreme (minimum or maximum) value of p(x) if p'(x) = 0

p'(x) = 2x-6 . Therefore 2x - 6 = 0 . Thus

x = 3 and the 2nd number is (-3) and the product is (-9).

7. Thanks so much for your explanations earboth--all make sense! On the last one though, isn't there another way to show it without taking derivative?

8. Originally Posted by sarahh
Thanks so much for your explanations earboth--all make sense! On the last one though, isn't there another way to show it without taking derivative?
Hi,

as you may have noticed the equation of p represents a parabola opening up. So the minimum of p is at it's vertex.
Probably you know to calculate the vertex of a parabola by using the completing-the-square-method.

9. Ahh that's right of course!! Thanks so much earboth!