Question

find the sum of all term between 98 and 509 divisible by 10​

Answer 1

Answer:

⌈10014⌉=8⌈10014⌉=8

therefor 8×14=1128×14=112 is the smallest number in between 100100-1,0001,000

divisible by 1414 and ⌊1,00014⌋=71⌊1,00014⌋=71 therefor 71×14=99471×14=994 is the biggest number n between 100100-1,0001,000 divisible by 1414. So the sum is 112+126+140+…+994=14×(8+9+10+…+71)=14×(0+1+2+3+…+71−(0+1+2+…+7))=14×(∑a=071a−∑b=07b)=14×(71×(71+1)2−7×(7+1)2)=…=14×2,528=35,392112+126+140+…+994=14×(8+9+10+…+71)=14×(0+1+2+3+…+71−(0+1+2+…+7))=14×(∑a=071a−∑b=07b)=14×(71×(71+1)2−7×(7+1)2)=…=14×2,528=35,392.

Answer 2

Hope it helps

Byeeeeeeeeeee