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Ujji

(10^25)-7 is divisible by a) 2 b)3 c)9 d) both 2 & 3

  • one year ago
  • one year ago

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  1. graydarl
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    b)3

    • one year ago
  2. graydarl
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    10^25=10000000000000000000000000 the number is not so important u know that the last digit will be 0 and is u substract 7 u will have tha last digit 3 si it is divisible by 3

    • one year ago
  3. Praja_01
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    10-7=3 therefore 10^25-7 is divisible by 3.

    • one year ago
  4. Praja_01
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    its not last digit but all preceeding digits are 9. therefore sum of 23 9s+3 is a multiple of 3.

    • one year ago
  5. graydarl
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    if u have 10 or 10^2 meaning 100 or 10^x it doesn't matter cause by substracting 7 u will have the last digit 3 and 3 is divisible only with 3

    • one year ago
  6. Praja_01
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    133 for instance is not divisible by 3.

    • one year ago
  7. graydarl
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    yup, u are right

    • one year ago
  8. shubhamsrg
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    anyways,alternately, we know, 10^n = 3q + 1 and 7 = 3p + 1 subtracting both, 10^n - 7 = 3(q-p) = 3r ==> divisible by 3

    • one year ago
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