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What least number must be subtracted from 1936 so that the resulting number when divided by 9,10 and 15 will leave in each case the same remainder 7 ?

137
236
339
430

Answer:- 1

Explanation:-

Solution:

\begin{aligned} LCM of (9, 10, 15) \\ = 3 \times 3 \times 10 \\ = 90 \\ \frac{1936}{90}, remainder  =  46 \end{aligned}

$\eqalign{ & {\text{LCM of }}\left( {9,10,15} \right) \cr & = 3 \times 3 \times 10 \cr & = 90 \cr & \frac{{1936}}{{90}},{\text{remainder = 46}} \cr}$ Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39

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Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39 ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} LCM of (9, 10, 15) \\ = 3 \times 3 \times 10 \\ = 90 \\ \frac{1936}{90}, remainder  =  46 \end{aligned}

$\eqalign{ & {\text{LCM of }}\left( {9,10,15} \right) \cr & = 3 \times 3 \times 10 \cr & = 90 \cr & \frac{{1936}}{{90}},{\text{remainder = 46}} \cr}$ Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39 ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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