Expresions(468) for Successive Integers

Here is a problem by J.A.H. Hunter on expressions for successive integers.

We're back to figures today: 4, 6 and 8 - all three, but only one of each. Using these, together with any regular mathematical signs but no other figures or words, you have to make up an expression for each number in turn working up from 'one.' You could express 38 as (46-8). That's an easy one.

You can use pluses, minuses, multiplication signs, division signs, brackets, powers, roots, decimals both regular and repeating, concatenation, factorials, and summation signs.

Use ROOT(n,m) as the nth root of m and SQRT as the square root. 6^SQRT(4)=36 and .ar=.aaaa... Use SUM(a,b,f(i)) as the summation function where a is the lower limit, b is the upper limit, and f(i) is a function of i. a and b must be expressions containing only some or all the integers(4,6,8) and if f(i) contains any numbers they must be selected from 4,6, and 8. Remember that 4, 6, and 8 must be used in the entire expression.

b

Using sums I can go further:

19 = SUM(6,8,i)-sqrt(4) 20 = 8+sqrt(4)*6 21 = (6+8)/sqrt(.4r) 22 = 6+sqrt(4)*8 23 = 8+6/.4 24 = 6*(8-4) 25 = (sqrt(.8-.6))^-4 26 = 6+8/.4 27 = 4*6/.8r 28 = sqrt(4)*(6+8) 29 = SUM(sqrt(4),8,i)-6 30 = 4*6/.8 31 = SUM(4,8,i-i/i)+6 32 = 8+4*6 33 = 6+4!/.8r 34 = (sqrt(sqrt(6)))^8-sqrt(4) 35 = (6+8)/.4 36 = 6^(8/4) 37 = SUM(4,6,i+i+i)-8 38 = 6+4*8 39 = SUM(sqrt(4)-6,8,i+i/i) 40 = (sqrt(sqrt(6)))^8+4

Please reply to ilan at cedara dot com

Nice work Ilan...

I was almost there myself - although I did figure that any even number can be created simply using:

SUM(SQRT(4),8-6,i+....+i)

with enough i's...

Good. However the expression for 39 has a problem as it includes indeterminate 0/0.

39=SUM(6,8,i+(i+i+i+i+i+i)/i+4*(i-i)) Anyone able to do 39 with less i's. An adaption of this method will work for any multiple of 3.

Here are a few more. 41=SUM(sqrt(4),8,i)+6 42=48-6 43=86/sqrt(4) 44=8*6-4 45=sqrt(6^4)/.8 46=8*6-sqrt(4)

The first even number that needs the SUM function is 74. However 47 needs the SUM function and is not a multiple of three.

[Problem definition and solutions from Ilan Mayer snipped...]

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39 = SUM(-SQRT(2),8,i) + 6

Andreas

SImplifying this for discussion, we have: n=sum (2,2,ni); n even

Unless I misread what you wrote, shouldn't this work for all numbers, not just even ones?

Very good. I am sure you meant 39 = SUM(-SQRT(4),8,i) + 6

No simplifying gets 2*n=sum(2,2,ni); n any integer

so it only works for for even numbers.

I just realized the error I made- I misread the notation of summation in this problem. You are right, it only works for even n.

As a side note, all multiples of 5 can be obtained by:

OK - I am really stretching the line here:

Multiples of 3: 3k = SUM(-4,6,ni)-8n

7 anyone?

No. While you can have as many i's as needed you can have only one 8 according to the puzzle. Actually I have already hinted how to get multiples of 3. Here it is explicitly:

3n = SUM(6,8,i+(n-7)*i/i+4*(i-i))