Math, asked by Limadagr5083, 1 year ago

Determine the sum of all possible positive 8ntegers the product of which digits equal to n^2-15n-27

Answers

Answered by krishtiwari07
0

Answer:

Product of digits = n2 – 15n – 27 = n(n – 15) – 27 Note that if n is a more than 2-digit number, say 3-digit number, then product has to be ≤ 9 × 9 × 9 = 729 but (n(n – 15) – 27) is more than 729 (in fact it a more than 3-digit numbers for any 3-digit n). Hence, n can be either one-digit or 2-digit. If n is 1-digit then n2 – 15n – 27 = n Read more on Sarthaks.com - https://www.sarthaks.com/333463/determine-the-sum-of-all-possible-positive-integers-the-product-of-whose-digits-equals-15n

Answered by UTM
0

Answer:

Taking n=1

->(-41)

Taking n=2

->(-53)

Taking n=3

->(-63)

Taking n=4

->(-71)

Taking n=5

->(-77)

Taking n=6

->(-81)

Taking n=7

->(-83)

Taking n=8

->(-83)

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