So, on to a more realistic solution, one that complies with the spirit rather than the letter of the rules.

My first instinct was to:

1-identify the first even number

2-Loop stepping by 2 adding the values until reach the top value.

I ended up with 100 loops. That's embarrassing! a hundred loop to solve a simple puzzle like this? There got to be a better way.

There is..

One that uses a 20 loops or few more.

Here it is:

P.S.: To best read the introduction in case your screen rsolution is different from mine, copy/paste it into Word and read it in Word.

Code:

'Introduction: 'The object is to sum all even values between 2 given numbers 'Since these 2 numbers are provided in the puzzle, meaning I could hard-code these values which makes everything a lot simpler and faster but just in case the intent was to solve for any numbers not just the ones provided in the puzzle, the script is written so that you can substitute these number at will and get the correct result. ' 'Let me first start with the logic behind the script. 'To add all even values between 2 given numbers, you identify the first even number, then add 2 to it and add the new number to the previous one and so on.. 'That's the standard way of thinking.. by doing that you will have to loop as many times as there are even values.. 'That's too many loops and too expensive.. ' 'Let's look at the puzzle from a different perspective.. ' 'In the example provided, the first even number is 102. 'Let's call this number a base. 'Every subsequent number is 2 points more than the one before.. 'Let's get away from this line of thinking. 'A different way to think about is: 'You have a base number, then 'Every subsequent number has one more 2 than the one before, i.e.: 'The base number is 102 'The one after has one more 2 than the base: 104 'The one after has two more 2's than the base: 106 'The one after has three more 2's than the base: 108 'and so on.. ' 'Now if I strip the base of its 2, then we could re-state the above as follows: 'Instead of the base, I am going to have new figure, will call it the "denominator" which is the base stripped off its 2, i.e. 102-2=100 ' 'Now the first even value (the base, 102) has one 2 more than the denominator and 'The second even value (104) has two 2's more than the denominator 'The third even value (106) has three 2's more than the denominator 'and so on.. ' 'so now I could deal with 2 sets of values: 'The first set, we will call it the "Denominator-cluster" is basically made up of the even values stripped off their 2's. meaning, they are all 100 ' 'The second set, we will call it the "Two-cluster" is made up of those 2's that are above and beyond the denominator. This is a series made up as one 2, two 2's, three 2's... ten 2's. ' 'Now I am going to break up all the even values between 101 and 302 into 10 clusters. 'As you can see, the "Denominator-clusters" will each have 10 equal values. the first cluster will have 10 equal values each of 100. If I loop through these clusters and step increment each cluster by 20 (10 2's), the second cluster will have 10 equal values of 120 each, the third cluster will have 10 equal values of 140 each.. and so on. ' 'To this cluster, I need to add the cluster of 2's. I need to calculate this cluster only once and add it to each one of the "denominator-clusters" and then add all the clusters together which gives me the final total sum. ' 'To add the cluster of 2's, I could hard-code that: '2+4+6+8...+20=110 'But once again, if the intent of the puzzle is to be able to substitute for other values, I will use a loop to add these incrementing 2's, which means I will loop 10 times to calculate the cluster of 2's then loop another 10 times to calculate the "denominator-clusters" and in each loop I am adding each one of the to the previous total plus the sum of the cluster of 2's so at the end of the last loop. After that I will find the orphan values, the ones that were not included in the clusters and add those to the total which gives me the grand total. 'Now why did I choose 10 clusters? 'That produces the "optimal" number of loops in this scenario: 'If you increase the size of the cluster to reduce their number and thus the number of looping through them, you will increase the number of 2's and the number of their loops. 'How to arrive at the optimal cluster size? 'That's for another day another post, but regardless, any number of clusters will beat the standard looping method.

Code:

'For further explanations, please read the Introduction section option strict dim bottom_num as n dim top_num as n dim denominator as n dim span as n 'the span between the bottom and top numbers dim values_num as n dim i as n 'to be used as a counter dim twos as n'this will add the 2's that are above and beyond the denominator values dim vtotal as n=0 dim trail_num as n'number of orphan values not in the clusters dim trail_values as n'the actual value of the orphan dim loop_num as n'how many times we looped so far dim last_even as n'this is the value of top-most even number in the series bottom_num=101 top_num=302 span=top_num-bottom_num values_num=int(span/2)+1 'This is how many even number in the span between the bottom and top numbers. The 1 is added to make it all inclusive, i.e. including the bottom as well as the top numbers trail_num=mod(values_num,10) denominator=bottom_num-if(mod(bottom_num,2)=0,0,1) 'First will add the "Two-cluster" for i=1 to 10 twos=twos+i*2 next i loop_num=i 'Now we will iterate over the "denominator-clusterss" , each cluster will have a denomiator value of 20 more the one before for i=1 to 10 vtotal=vtotal+(denominator+20*(i-1))*10+twos next i loop_num=loop_num+i 'Now we will add the trailing numbers, those that were not included in the 10 clusters 'Find the last even number: last_even=if(mod(top_num,2)=0,top_num,top_num-1) for i= 1 to trail_num step-1 trail_values=trail_values+last_even last_even=last_even-2 next loop_num=loop_num+1 vtotal=vtotal+trail_values msgbox("The sum total of all even numbers is: "+vtotal+" "+"and the number of loops is: "+loop_num)

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