allery12, 18 କିମ୍ବା 28 ଲେଖାଏଁ ପ୍ୟାକ୍ କଲେ 5ଟି ସାମଗ୍ରୀ ବାକି ରହେ। 1000 ରୁ ଅଧିକ ସର୍ବନିମ୍ନ ମଜୁତ ସଂଖ୍ୟା କେତେ?
A1005
B1013
C1025
D1265
Answer & Solution
Correct answer: B. 1013
LCM(12, 18, 28) = 252, ତେଣୁ 1000 ରୁ ଅଧିକ ସର୍ବନିମ୍ନ ଯୋଗ୍ୟ ମଜୁତ 4 × 252 + 5 = 1013।
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