How to generate number 98 using all 4 digits 1, 8, 4, 6 just once

[cris0972]

and any math operation such as: +, -, *, :, sqrt, !, (), decimal point (.8, .4). Also digits can be combined to generate a new number: 1&4: 14, 8&6: 86.
Example:

62 = 86 - 4! x 1
97 = (4! - 8) x 6 +1



Thanks !!

 

[JeffM]

and any math operation such as: +, -, *, :, sqrt, !, (), decimal point (.8, .4). Also digits can be combined to generate a new number: 1&4: 14, 8&6: 86.
Example:

62 = 86 - 4! x 1
97 = (4! - 8) x 6 +1



Thanks !!

There may be an infinite number of ways to solve this problem. Consequently, I do not think such problems teach much that is useful.

One possible answer is (4 + 6 - 1) & 8 giving 9 & 8 = 98.

 

[cris0972]

6! / 8 + 4/1

Denis, 6!/8 = 90
4/1 = 4
So the end results would be 94, and not 98, no ?

 

[cris0972]

There may be an infinite number of ways to solve this problem. Consequently, I do not think such problems teach much that is useful.

One possible answer is (4 + 6 - 1) & 8 giving 9 & 8 = 98.

Sorry, maybe I was not clear ... you can combine the dights to generate a new number, but you can't "link" 2 digits to get 98

JeffM, out of those infinite numbres, can you just suggest one?
Actually the exercise was to generate all numbers 1-100 and as far as I can tell it was a "cool" exercise for the 4th graders.
Thx

 

[cris0972]

My bad; need another coffee!
Trying again...symbol ^ (to the power) not allowed?

^ - allowed; like in 4^1 = 4 or 1^8 = 1

Thanks a lot !!

 

[lookagain]

and any math operation such as: +, -, *, :, sqrt, !, (), decimal point (.8, .4).

Also digits can be combined to generate a new number: 1&4: 14, 8&6: 86.
Example:

62 = 86 - 4! x 1
97 = (4! - 8) x 6 +1

Here is one of the ways:



(8*6 + 1)(sqrt(4)) = 98


or, in Latex, as


$\displaystyle (8*6 \ + \ 1)\sqrt{4} \ = 98$

 

[cris0972]

Hats off !! Thank you both.
Denis - I'll try 2 operators for 35 but don't promise anything

 

[cris0972]

Challenge to you Cris: with your puzzle, get 35 using only 2 operations.

81 - 46 = 35 will that work ?
and it's one operator

 

[Deleted member 4993]

Shucks! 2 operators, please :???:

What about the "=" sign!!

 

[lookagain]

Nice one lookagain!Mine's better:8 / .1 + 4! - 6 = 98 ...no yukky brackets :roll:

Oh, you mean "yucky" brackets.

$\displaystyle (8*6 \ + \ 1)\sqrt{4} \ = 98$

So, you're going to force a comparison between yours and mine about being "better,"

which is subjective anyway.

To the left of the equals sign in our equations, each of us used nine characters not including spaces.

I used (in no particular order) the 1) multiplication sign, 2) the addition sign, 3) the square root symbol,

and 4) the one pair of parentheses, which gets me a twofer for the grouping and another for the

multiplication by the square root of four.

Contrast that with yours:

1) a division sign, 2) a decimal point, 3) an addition sign, 4) a factorial symbol, and 5) a subtraction sign

And two places in yours, a non-digit immediately follows another non-digit, that is, 1) and then 2), and then later

4) and then 5).

Whereas for me, just the square root symbol follows the close parenthesis.

Based on those criteria, yours squeaks out as being more complicated, but I won't ascribe the subjective word "better" to mine.

 

[guitarguy]

Interesting problem. If you can use square root can you also use squared, if so I got:

((8-1)^2)(6-4)=98

 

[Deleted member 4993]

Interesting problem. If you can use square root can you also use squared, if so I got:

((8-1)^2)(6-4)=98

Unfortunately, square-root works because it has its own sign (technically though it should be written as $\displaystyle \displaystyle{\sqrt[2]{841}}$) - and for 'square' you have to write the number 2 - and that is not included in the allowed numbers. Lookagain and Denis cheated!!

 

[lookagain]

Unfortunately, square-root works because it has its own sign (technically though it should be written as $\displaystyle \displaystyle{\sqrt[2]{841}}$) - and for 'square' you have to write the number 2 - and that is not included in the allowed numbers. Lookagain and Denis cheated!!

8 / .1 + 4! - 6 = 98 ...

If the square root symbol shouldn't be allowed because it implies the digit 2, then the factorial symbol, as in Denis's solution above,also shouldn't be allowed, because it implies multiplication by these: 4! = 4*3*2*1.

I have no problem forbidding the use of the square root symbol (or its equivalent in words) for these puzzles.

 

[cris0972]

Unfortunately, square-root works because it has its own sign (technically though it should be written as $\displaystyle \displaystyle{\sqrt[2]{841}}$) - and for 'square' you have to write the number 2 - and that is not included in the allowed numbers. Lookagain and Denis cheated!!

Got your point, however, for the purpose of this exercise, square-root was allowed. I guess because it's using the square-root symbol.